5.7: Surface Integrals
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Learning Objectives
- Find the parametric representations of a cylinder, a cone, and a sphere.
- Describe the surface integral of a scalar-valued function over a parametric surface.
- Use a surface integral to calculate the area of a given surface.
- Explain the meaning of an oriented surface, giving an example.
- Describe the surface integral of a vector field.
- Use surface integrals to solve applied problems.
We have seen that a line integral is an integral over a path in a plane or in space. However, if we wish to integrate over a surface (a two-dimensional object) rather than a path (a one-dimensional object) in space, then we need a new kind of integral that can handle integration over objects in higher dimensions. We can extend the concept of a line integral to a surface integral to allow us to perform this integration.
Surface integrals are important for the same reasons that line integrals are important. They have many applications to physics and engineering, and they allow us to develop higher dimensional versions of the Fundamental Theorem of Calculus . In particular, surface integrals allow us to generalize Green’s theorem to higher dimensions, and they appear in some important theorems we discuss in later sections.
Parametric Surfaces
A surface integral is similar to a line integral, except the integration is done over a surface rather than a path. In this sense, surface integrals expand on our study of line integrals. Just as with line integrals, there are two kinds of surface integrals: a surface integral of a scalar-valued function and a surface integral of a vector field.
However, before we can integrate over a surface, we need to consider the surface itself. Recall that to calculate a scalar or vector line integral over curve \(C\), we first need to parameterize \(C\). In a similar way, to calculate a surface integral over surface \(S\), we need to parameterize \(S\). That is, we need a working concept of a parameterized surface (or a parametric surface ), in the same way that we already have a concept of a parameterized curve.
A parameterized surface is given by a description of the form
\[\vecs{r}(u,v) = \langle x (u,v), \, y(u,v), \, z(u,v)\rangle. \nonumber \]
Notice that this parameterization involves two parameters, \(u\) and \(v\), because a surface is two-dimensional, and therefore two variables are needed to trace out the surface. The parameters \(u\) and \(v\) vary over a region called the parameter domain, or parameter space —the set of points in the \(uv\)-plane that can be substituted into \(\vecs r\).  The entire surface is created by making all possible choices of \(u\) and \(v\) over the parameter domain.
Definition: Parameter Domain
Given a parameterization of surface
\[\vecs{r}(u,v) = \langle x (u,v), \, y(u,v), \, z(u,v)\rangle. \nonumber \]
the parameter domain of the parameterization is the set of points in the \(uv\)-plane that can be substituted into \(\vecs r\).
 Since the parameter domain is all of \(\mathbb{R}^2\), we can choose any value for u and v and plot the corresponding point. If \(u = v = 0\), then \(\vecs r(0,0) = \langle 1,0,0 \rangle\), so point (1, 0, 0) is on \(S\). Similarly, points \(\vecs r(\pi, 2) = (-1,0,2)\) and \(\vecs r \left(\dfrac{\pi}{2}, 4\right) = (0,1,4)\) are on \(S\).
Although plotting points may give us an idea of the shape of the surface, we usually need quite a few points to see the shape. Since it is time-consuming to plot dozens or hundreds of points, we use another strategy. To visualize \(S\), we visualize two families of curves that lie on \(S\) . In the first family of curves we hold \(u\) constant; in the second family of curves we hold \(v\) constant. This allows us to build a “skeleton” of the surface, thereby getting an idea of its shape.
- Suppose that \(u\) is a constant \(K\). Then the curve traced out by the parameterization is \(\langle \cos K, \, \sin K, \, v \rangle \), which gives a vertical line that goes through point \((\cos K, \sin K, v \rangle\) in the \(xy\)-plane.
- Suppose that \(v\) is a constant \(K\). Then the curve traced out by the parameterization is \(\langle \cos u, \, \sin u, \, K \rangle \), which gives a circle in plane \(z = K\) with radius 1 and center \((0, 0, K)\).
If \(u\) is held constant, then we get vertical lines; if \(v\) is held constant, then we get circles of radius 1 centered around the vertical line that goes through the origin. Therefore the surface traced out by the parameterization is cylinder \(x^2 + y^2 = 1\) (Figure \(\PageIndex{1}\)).
Notice that if \(x = \cos u\) and \(y = \sin u\), then \(x^2 + y^2 = 1\), so points from S do indeed lie on the cylinder. Conversely, each point on the cylinder is contained in some circle \(\langle \cos u, \, \sin u, \, k \rangle \) for some \(k\), and therefore each point on the cylinder is contained in the parameterized surface (Figure \(\PageIndex{2}\)).
Analysis
Notice that if we change the parameter domain, we could get a different surface. For example, if we restricted the domain to \(0 \leq u \leq \pi, \, -\infty < v < 6\), then the surface would be a half-cylinder of height 6.
Exercise \(\PageIndex{1}\)
Describe the surface with parameterization
\[\vecs{r} (u,v) = \langle 2 \, \cos u, \, 2 \, \sin u, \, v \rangle, \, 0 \leq u \leq 2\pi, \, -\infty < v < \infty \nonumber \]
- Hint
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Hold \(u\) and \(v\) constant, and see what kind of curves result.
- Answer
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Cylinder \(x^2 + y^2 = 4\)
Example \(\PageIndex{3}\): Finding a Parameterization
Give a parameterization of the cone \(x^2 + y^2 = z^2\) lying on or above the plane \(z = -2\).
Solution
The horizontal cross-section of the cone at height \(z = u\) is circle \(x^2 + y^2 = u^2\). Therefore, a point on the cone at height \(u\) has coordinates \((u \, \cos v, \, u \, \sin v, \, u)\) for angle \(v\).  Since we are not interested in the entire cone, only the portion on or above plane \(z = -2\), the parameter domain is given by \(-2 < u < \infty, \, 0 \leq v < 2\pi\) (Figure \(\PageIndex{4}\)).
Exercise \(\PageIndex{3}\)
Give a parameterization for the portion of cone \(x^2 + y^2 = z^2\) lying in the first octant.
- Hint
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Consider the parameter domain for this surface.
- Answer
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\(\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, u \rangle, \, 0 < u < \infty, \, 0 \leq v < \dfrac{\pi}{2}\)
We have discussed parameterizations of various surfaces, but two important types of surfaces need a separate discussion: spheres and graphs of two-variable functions. To parameterize a sphere, it is easiest to use spherical coordinates. The sphere of radius \(\rho\) centered at the origin is given by the parameterization
\(\vecs r(\phi,\theta) = \langle \rho \, \cos \theta \, \sin \phi, \, \rho \, \sin \theta \, \sin \phi, \, \rho \, \cos \phi \rangle, \, 0 \leq \theta \leq 2\pi, \, 0 \leq \phi \leq \pi.\)
The idea of this parameterization is that as \(\phi\) sweeps downward from the positive \(z\)-axis, a circle of radius \(\rho \, \sin \phi\) is traced out by letting \(\theta\) run from 0 to \(2\pi\). To see this, let \(\phi\) be fixed. Then
\[\begin{align*} x^2 + y^2 &= (\rho \, \cos \theta \, \sin \phi)^2 + (\rho \, \sin \theta \, \sin \phi)^2 \\[4pt]
&= \rho^2 \sin^2 \phi (\cos^2 \theta + \sin^2 \theta) \\[4pt]
&= \rho^2 \, \sin^2 \phi \\[4pt]
&= (\rho \, \sin \phi)^2. \end{align*}\]
This results in the desired circle (Figure \(\PageIndex{5}\)).
Finally, to parameterize the graph of a two-variable function, we first let \(z = f(x,y)\) be a function of two variables. The simplest parameterization of the graph of \(f\) is \(\vecs r(x,y) = \langle x,y,f(x,y) \rangle\), where \(x\) and \(y\) vary over the domain of \(f\) (Figure \(\PageIndex{6}\)). For example, the graph of \(f(x,y) = x^2 y\) can be parameterized by \(\vecs r(x,y) = \langle x,y,x^2y \rangle\), where the parameters \(x\) and \(y\) vary over the domain of \(f\). If we only care about a piece of the graph of \(f\) - say, the piece of the graph over rectangle \([ 1,3] \times [2,5]\) - then we can restrict the parameter domain to give this piece of the surface:
\[\vecs r(x,y) = \langle x,y,x^2y \rangle, \, 1 \leq x \leq 3, \, 2 \leq y \leq 5. \nonumber \]
Similarly, if \(S\) is a surface given by equation \(x = g(y,z)\) or equation \(y = h(x,z)\), then a parameterization of \(S\) is \(\vecs r(y,z) = \langle g(y,z), \, y,z\rangle\) or \(\vecs r(x,z) = \langle x,h(x,z), z\rangle\), respectively. For example, the graph of paraboloid \(2y = x^2 + z^2\) can be parameterized by \(\vecs r(x,y) = \left\langle x, \dfrac{x^2+z^2}{2}, z \right\rangle, \, 0 \leq x < \infty, \, 0 \leq z < \infty\). Notice that we do not need to vary over the entire domain of \(y\) because \(x\) and \(z\) are squared.
Let’s now generalize the notions of smoothness and regularity to a parametric surface. Recall that curve parameterization \(\vecs r(t), \, a \leq t \leq b\) is regular (or smooth ) if \(\vecs r'(t) \neq \vecs 0\) for all \(t\) in \([a,b]\). For a curve, this condition ensures that the image of \(\vecs r\) really is a curve, and not just a point. For example, consider curve parameterization \(\vecs r(t) = \langle 1,2\rangle, \, 0 \leq t \leq 5\). The image of this parameterization is simply point \((1,2)\), which is not a curve. Notice also that \(\vecs r'(t) = \vecs 0\). The fact that the derivative is the zero vector indicates we are not actually looking at a curve.
Analogously, we would like a notion of regularity (or smoothness) for surfaces so that a surface parameterization really does trace out a surface. To motivate the definition of regularity of a surface parameterization, consider the parameterization
\[\vecs r(u,v) = \langle 0, \, \cos v, \, 1 \rangle, \, 0 \leq u \leq 1, \, 0 \leq v \leq \pi. \nonumber \]
Although this parameterization appears to be the parameterization of a surface, notice that the image is actually a line (Figure \(\PageIndex{7}\)). How could we avoid parameterizations such as this? Parameterizations that do not give an actual surface? Notice that \(\vecs r_u = \langle 0,0,0 \rangle\) and \(\vecs r_v = \langle 0, -\sin v, 0\rangle\), and the corresponding cross product is zero. The analog of the condition \(\vecs r'(t) = \vecs 0\) is that \(\vecs r_u \times \vecs r_v\) is not zero for point \((u,v)\) in the parameter domain, which is a regular parameterization.
Definition: Regular Parameterization
Parameterization \(\vecs r(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle\) is a regular parameterization if \(\vecs r_u \times \vecs r_v\) is not zero for point \((u,v)\) in the parameter domain.
If parameterization \(\vec{r}\) is regular, then the image of \(\vec{r}\) is a two-dimensional object, as a surface should be. Throughout this chapter, parameterizations \(\vecs r(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle\)are assumed to be regular.
Recall that curve parameterization \(\vecs r(t), \, a \leq t \leq b\) is smooth if \(\vecs r'(t)\) is continuous and \(\vecs r'(t) \neq \vecs 0\) for all \(t\) in \([a,b]\). Informally, a curve parameterization is smooth if the resulting curve has no sharp corners. The definition of a smooth surface parameterization is similar. Informally, a surface parameterization is smooth if the resulting surface has no sharp corners.
Definition: Smooth Surfaces
A surface parameterization \(\vecs r(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle\) is smooth if vector \(\vecs r_u \times \vecs r_v\) is not zero for any choice of \(u\) and \(v\) in the parameter domain.
A surface may also be piecewise smooth if it has smooth faces but also has locations where the directional derivatives do not exist.
Example \(\PageIndex{4}\): Identifying Smooth and Nonsmooth Surfaces
Which of the figures in Figure \(\PageIndex{8}\) is smooth?
Solution
The surface in Figure \(\PageIndex{8a}\) can be parameterized by
\[\vecs r(u,v) = \langle (2 + \cos v) \cos u, \, (2 + \cos v) \sin u, \, \sin v \rangle, \, 0 \leq u < 2\pi, \, 0 \leq v < 2\pi \nonumber \]
(we can use technology to verify).  If \(v = 0\) or \(v = \pi\), then the only choices for \(u\) that make the \(\mathbf{\hat{j}}\) component zero are \(u = 0\) or \(u = \pi\). But, these choices of \(u\) do not make the \(\mathbf{\hat{i}}\) component zero. Therefore, \(\vecs r_u \times \vecs r_v\) is not zero for any choice of \(u\) and \(v\) in the parameter domain, and the parameterization is smooth. Notice that the corresponding surface has no sharp corners.
In the pyramid in Figure \(\PageIndex{8b}\), the sharpness of the corners ensures that directional derivatives do not exist at those locations. Therefore, the pyramid has no smooth parameterization. However, the pyramid consists of four smooth faces, and thus this surface is piecewise smooth.
Exercise \(\PageIndex{4}\)
Is the surface parameterization \(\vecs r(u,v) = \langle u^{2v}, v + 1, \, \sin u \rangle, \, 0 \leq u \leq 2, \, 0 \leq v \leq 3\) smooth?
- Hint
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Investigate the cross product \(\vecs r_u \times \vecs r_v\).
- Answer
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Yes
Surface Area of a Parametric Surface
Our goal is to define a surface integral, and as a first step we have examined how to parameterize a surface. The second step is to define the surface area of a parametric surface. The notation needed to develop this definition is used throughout the rest of this chapter.
Let \(S\) be a surface with parameterization \(\vecs r(u,v) = \langle x(u,v), \, y(u,v), \, z(u,v) \rangle\) over some parameter domain \(D\). We assume here and throughout that the surface parameterization \(\vecs r(u,v) = \langle x(u,v), \, y(u,v), \, z(u,v) \rangle\) is continuously differentiable—meaning, each component function has continuous partial derivatives. Assume for the sake of simplicity that \(D\) is a rectangle (although the following material can be extended to handle nonrectangular parameter domains). Divide rectangle \(D\) into subrectangles \(D_{ij}\) with horizontal width \(\Delta u\) and vertical length \(\Delta v\). Suppose that i ranges from 1 to m and j ranges from 1 to n so that \(D\) is subdivided into mn rectangles.  Choose point \(P_{ij}\) in each piece \(S_{ij}\). Point \(P_{ij}\) corresponds to point \((u_i, v_j)\) in the parameter domain.
Note that we can form a grid with lines that are parallel to the \(u\)-axis and the \(v\)-axis in the \(uv\)-plane. These grid lines correspond to a set of grid curves on surface \(S\) that is parameterized by \(\vecs r(u,v)\). Without loss of generality, we assume that \(P_{ij}\) is located at the corner of two grid curves, as in Figure \(\PageIndex{9}\). If we think of \(\vecs r\) as a mapping from the \(uv\)-plane to \(\mathbb{R}^3\), the grid curves are the image of the grid lines under \(\vecs r\). To be precise, consider the grid lines that go through point \((u_i, v_j)\). One line is given by \(x = u_i, \, y = v\); the other is given by \(x = u, \, y = v_j\). In the first grid line, the horizontal component is held constant, yielding a vertical line through \((u_i, v_j)\). In the second grid line, the vertical component is held constant, yielding a horizontal line through \((u_i, v_j)\). The corresponding grid curves are \(\vecs r(u_i, v)\) and \((u, v_j)\) and these curves intersect at point \(P_{ij}\).
Now consider the vectors that are tangent to these grid curves. For grid curve \(\vecs r(u_i,v)\), the tangent vector at \(P_{ij}\) is
\[\vecs t_v (P_{ij}) = \vecs r_v (u_i,v_j) = \langle x_v (u_i,v_j), \, y_v(u_i,v_j), \, z_v (u_i,v_j) \rangle. \nonumber \]
For grid curve \(\vecs r(u, v_j)\), the tangent vector at \(P_{ij}\) is
\[\vecs t_u (P_{ij}) = \vecs r_u (u_i,v_j) = \langle x_u (u_i,v_j), \, y_u(u_i,v_j), \, z_u (u_i,v_j) \rangle. \nonumber \]
If vector \(\vecs N = \vecs t_u (P_{ij}) \times \vecs t_v (P_{ij})\) exists and is not zero, then the tangent plane at \(P_{ij}\) exists (Figure \(\PageIndex{10}\)). If piece \(S_{ij}\) is small enough, then the tangent plane at point \(P_{ij}\) is a good approximation of piece \(S_{ij}\).
The tangent plane at \(P_{ij}\) contains vectors \(\vecs t_u(P_{ij})\) and \(\vecs t_v(P_{ij})\) and therefore the parallelogram spanned by \(\vecs t_u(P_{ij})\) and \(\vecs t_v(P_{ij})\) is in the tangent plane. Since the original rectangle in the \(uv\)-plane corresponding to \(S_{ij}\) has width \(\Delta u\) and length \(\Delta v\), the parallelogram that we use to approximate \(S_{ij}\) is the parallelogram spanned by \(\Delta u \vecs t_u(P_{ij})\) and \(\Delta v \vecs t_v(P_{ij})\). In other words, we scale the tangent vectors by the constants \(\Delta u\) and \(\Delta v\) to match the scale of the original division of rectangles in the parameter domain. Therefore, the area of the parallelogram used to approximate the area of \(S_{ij}\) is
\[\Delta S_{ij} \approx ||(\Delta u \vecs t_u (P_{ij})) \times (\Delta v \vecs t_v (P_{ij})) || = ||\vecs t_u (P_{ij}) \times \vecs t_v (P_{ij}) || \Delta u \,\Delta v. \nonumber \]
Varying point \(P_{ij}\) over all pieces \(S_{ij}\) and the previous approximation leads to the following definition of surface area of a parametric surface (Figure \(\PageIndex{11}\)).
Definition: Smooth Parameterization of Surface
Let \(\vecs r(u,v) = \langle x(u,v), \, y(u,v), \, z(u,v) \rangle\) with parameter domain \(D\) be a smooth parameterization of surface \(S\). Furthermore, assume that \(S\) is traced out only once as \((u,v)\) varies over \(D\). The surface area of \(S\) is
\[\iint_D ||\vecs t_u \times \vecs t_v || \,dA, \label{equation1} \]
where \(\vecs t_u = \left\langle \dfrac{\partial x}{\partial u},\, \dfrac{\partial y}{\partial u},\, \dfrac{\partial z}{\partial u} \right\rangle\)
and
\[\vecs t_v = \left\langle \dfrac{\partial x}{\partial v},\, \dfrac{\partial y}{\partial v},\, \dfrac{\partial z}{\partial v} \right\rangle. \nonumber \]
Example \(\PageIndex{5}\): Calculating Surface Area
Calculate the lateral surface area (the area of the “side,” not including the base) of the right circular cone with height \(h\) and radius \(r\).
Solution
Before calculating the surface area of this cone using Equation \ref{equation1}, we need a parameterization. We assume this cone is in \(\mathbb{R}^3\) with its vertex at the origin (Figure \(\PageIndex{12}\)). To obtain a parameterization, let \(\alpha\) be the angle that is swept out by starting at the positive z -axis and ending at the cone, and let \(k = \tan \alpha\). For a height value \(v\) with \(0 \leq v \leq h\), the radius of the circle formed by intersecting the cone with plane \(z = v\) is \(kv\). Therefore, a parameterization of this cone is
\[\vecs s(u,v) = \langle kv \, \cos u, \, kv \, \sin u, \, v \rangle, \, 0 \leq u < 2\pi, \, 0 \leq v \leq h. \nonumber \]
The idea behind this parameterization is that for a fixed \(v\)-value, the circle swept out by letting \(u\) vary is the circle at height \(v\) and radius \(kv\). As \(v\) increases, the parameterization sweeps out a “stack” of circles, resulting in the desired cone.
With a parameterization in hand, we can calculate the surface area of the cone using Equation \ref{equation1}. The tangent vectors are \(\vecs t_u = \langle - kv \, \sin u, \, kv \, \cos u, \, 0 \rangle\) and \(\vecs t_v = \langle k \, \cos u, \, k \, \sin u, \, 1 \rangle\).
AnalysisThe surface area of a right circular cone with radius \(r\) and height \(h\) is usually given as \(\pi r^2 + \pi r \sqrt{h^2 + r^2}\). The reason for this is that the circular base is included as part of the cone, and therefore the area of the base \(\pi r^2\) is added to the lateral surface area \(\pi r \sqrt{h^2 + r^2}\) that we found.

- Hint
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Use Equation \ref{equation1}.
- Answer
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\(≈ 43.02\)
Example \(\PageIndex{6}\): Calculating Surface Area
Show that the surface area of the sphere \(x^2 + y^2 + z^2 = r^2\) is \(4 \pi r^2\).
Solution
The sphere has parameterization
\(r \, \cos \theta \, \sin \phi, \, r \, \sin \theta \, \sin \phi, \, r \, \cos \phi \rangle, \, 0 \leq \theta < 2\pi, \, 0 \leq \phi \leq \pi.\)

 The surface area of the sphere is
\[\int_0^{2\pi} \int_0^{\pi} r^2 \sin \phi \, d\phi \,d\theta = r^2 \int_0^{2\pi} 2 \, d\theta = 4\pi r^2. \nonumber \]
We have derived the familiar formula for the surface area of a sphere using surface integrals.
Exercise \(\PageIndex{6}\)
Show that the surface area of cylinder \(x^2 + y^2 = r^2, \, 0 \leq z \leq h\) is \(2\pi rh\). Notice that this cylinder does not include the top and bottom circles.
- Hint
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Use the standard parameterization of a cylinder and follow the previous example.
- Answer
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With the standard parameterization of a cylinder, Equation \ref{equation1} shows that the surface area is \(2 \pi rh\).
In addition to parameterizing surfaces given by equations or standard geometric shapes such as cones and spheres, we can also parameterize surfaces of revolution. Therefore, we can calculate the surface area of a surface of revolution by using the same techniques. Let \(y = f(x) \geq 0\) be a positive single-variable function on the domain \(a \leq x \leq b\) and let \(S\) be the surface obtained by rotating \(f\) about the \(x\)-axis (Figure \(\PageIndex{13}\)). Let \(\theta\) be the angle of rotation. Then, \(S\) can be parameterized with parameters \(x\) and \(\theta\) by
\[\vecs r(x, \theta) = \langle x, f(x) \, \cos \theta, \, f(x) \sin \theta \rangle, \, a \leq x \leq b, \, 0 \leq x \leq 2\pi. \nonumber \]
Example \(\PageIndex{7}\): Calculating Surface Area
Find the area of the surface of revolution obtained by rotating \(y = x^2, \, 0 \leq x \leq b\) about the \(x\)-axis (Figure \(\PageIndex{14}\)).
Solution
This surface has parameterization \(\vecs r(x, \theta) = \langle x, \, x^2 \cos \theta, \, x^2 \sin \theta \rangle, \, 0 \leq x \leq b, \, 0 \leq x < 2\pi.\)
The tangent vectors are \( \vecs t_x = \langle 1, \, 2x \, \cos \theta, \, 2x \, \sin \theta \rangle\) and \(\vecs t_{\theta} = \langle 0, \, -x^2 \sin \theta, \, -x^2 \cos \theta \rangle\).
Therefore,
\[\begin{align*} \vecs t_x \times \vecs t_{\theta} &= \langle 2x^3 \cos^2 \theta + 2x^3 \sin^2 \theta, \, -x^2 \cos \theta, \, -x^2 \sin \theta \rangle \\[4pt] &= \langle 2x^3, \, -x^2 \cos \theta, \, -x^2 \sin \theta \rangle \end{align*}\]
and
\[\begin{align*} \vecs t_x \times \vecs t_{\theta} &= \sqrt{4x^6 + x^4\cos^2 \theta + x^4 \sin^2 \theta} \\[4pt] &= \sqrt{4x^6 + x^4} \\[4pt] &= x^2 \sqrt{4x^2 + 1} \end{align*}\]
The area of the surface of revolution is
\[\begin{align*} \int_0^b \int_0^{2\pi} x^2 \sqrt{4x^2 + 1} \, d\theta \,dx &= 2\pi \int_0^b x^2 \sqrt{4x^2 + 1} \,dx \\[4pt]
&= 2\pi \left[ \dfrac{1}{64} \left(2 \sqrt{4x^2 + 1} (8x^3 + x) \, \sinh^{-1} (2x)\right)\right]_0^b \\[4pt]
&= 2\pi \left[ \dfrac{1}{64} \left(2 \sqrt{4b^2 + 1} (8b^3 + b) \, \sinh^{-1} (2b) \right)\right]. \end{align*}\]
Exercise \(\PageIndex{7}\)
Use Equation \ref{equation1} to find the area of the surface of revolution obtained by rotating curve \(y = \sin x, \, 0 \leq x \leq \pi\) about the \(x\)-axis.
- Hint
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Use the parameterization of surfaces of revolution given before Example \(\PageIndex{7}\).
- Answer
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\(2\pi (\sqrt{2} + \sinh^{-1} (1))\)
Surface Integral of a Scalar-Valued Function
Now that we can parameterize surfaces and we can calculate their surface areas, we are able to define surface integrals. First, let’s look at the surface integral of a scalar-valued function. Informally, the surface integral of a scalar-valued function is an analog of a scalar line integral in one higher dimension. The domain of integration of a scalar line integral is a parameterized curve (a one-dimensional object); the domain of integration of a scalar surface integral is a parameterized surface (a two-dimensional object). Therefore, the definition of a surface integral follows the definition of a line integral quite closely. For scalar line integrals, we chopped the domain curve into tiny pieces, chose a point in each piece, computed the function at that point, and took a limit of the corresponding Riemann sum. For scalar surface integrals, we chop the domain region (no longer a curve) into tiny pieces and proceed in the same fashion.
Let \(S\) be a piecewise smooth surface with parameterization \(\vecs{r}(u,v) = \langle x(u,v), \, y(u,v), \, z(u,v) \rangle \) with parameter domain \(D\) and let \(f(x,y,z)\) be a function with a domain that contains \(S\) . For now, assume the parameter domain \(D\) is a rectangle, but we can extend the basic logic of how we proceed to any parameter domain (the choice of a rectangle is simply to make the notation more manageable). Divide rectangle \(D\) into subrectangles \(D_{ij}\) with horizontal width \(\Delta u\) and vertical length \(\Delta v\). Suppose that \(i\) ranges from \(1\) to \(m\) and \(j\) ranges from \(1\) to \(n\) so that \(D\) is subdivided into \(mn\) rectangles. This division of \(D\) into subrectangles gives a corresponding division of \(S\) into pieces \(S_{ij}\). Choose point \(P_{ij}\) in each piece \(S_{ij}\) evaluate \(P_{ij}\) at \(f\), and multiply by area \(S_{ij}\) to form the Riemann sum
\[\sum_{i=1}^m \sum_{j=1}^n f(P_{ij}) \, \Delta S_{ij}. \nonumber \]
To define a surface integral of a scalar-valued function, we let the areas of the pieces of \(S\) shrink to zero by taking a limit.
Definition: Surface Integral of a Scalar-Valued Function
The surface integral of a scalar-valued functi on of \(f\) over a piecewise smooth surface \(S\) is
\[\iint_S f(x,y,z) dA = \lim_{m,n\rightarrow \infty} \sum_{i=1}^m \sum_{j=1}^n f(P_{ij}) \Delta S_{ij}. \nonumber \]
Again, notice the similarities between this definition and the definition of a scalar line integral. In the definition of a line integral we chop a curve into pieces, evaluate a function at a point in each piece, and let the length of the pieces shrink to zero by taking the limit of the corresponding Riemann sum. In the definition of a surface integral, we chop a surface into pieces, evaluate a function at a point in each piece, and let the area of the pieces shrink to zero by taking the limit of the corresponding Riemann sum. Thus, a surface integral is similar to a line integral but in one higher dimension.
The definition of a scalar line integral can be extended to parameter domains that are not rectangles by using the same logic used earlier. The basic idea is to chop the parameter domain into small pieces, choose a sample point in each piece, and so on. The exact shape of each piece in the sample domain becomes irrelevant as the areas of the pieces shrink to zero.
Scalar surface integrals are difficult to compute from the definition, just as scalar line integrals are. To develop a method that makes surface integrals easier to compute, we approximate surface areas \(\Delta S_{ij}\) with small pieces of a tangent plane, just as we did in the previous subsection. Recall the definition of vectors \(\vecs t_u\) and \(\vecs t_v\):
\[\vecs t_u = \left\langle \dfrac{\partial x}{\partial u},\, \dfrac{\partial y}{\partial u},\, \dfrac{\partial z}{\partial u} \right\rangle\, \text{and} \, \vecs t_v = \left\langle \dfrac{\partial x}{\partial v},\, \dfrac{\partial y}{\partial v},\, \dfrac{\partial z}{\partial v} \right\rangle. \nonumber \]
From the material we have already studied, we know that
\[\Delta S_{ij} \approx ||\vecs t_u (P_{ij}) \times \vecs t_v (P_{ij})|| \,\Delta u \,\Delta v. \nonumber \]
Therefore,
\[\iint_S f(x,y,z) \,dS \approx \lim_{m,n\rightarrow\infty} \sum_{i=1}^m \sum_{j=1}^n f(P_{ij})|| \vecs t_u(P_{ij}) \times \vecs t_v(P_{ij}) ||\,\Delta u \,\Delta v. \nonumber \]
This approximation becomes arbitrarily close to \(\displaystyle \lim_{m,n\rightarrow\infty} \sum_{i=1}^m \sum_{j=1}^n f(P_{ij}) \Delta S_{ij}\) as we increase the number of pieces \(S_{ij}\) by letting \(m\) and \(n\) go to infinity. Therefore, we have the following equation to calculate scalar surface integrals :
\[\iint_S f(x,y,z)\,dS = \iint_D f(\vecs r(u,v)) ||\vecs t_u \times \vecs t_v||\,dA. \label{scalar surface integrals} \]
Equation \ref{scalar surface integrals} allows us to calculate a surface integral by transforming it into a double integral. This equation for surface integrals is analogous to the equation for line integrals:
\[\iint_C f(x,y,z)\,ds = \int_a^b f(\vecs r(t))||\vecs r'(t)||\,dt. \nonumber \]
In this case, vector \(\vecs t_u \times \vecs t_v\) is perpendicular to the surface, whereas vector \(\vecs r'(t)\) is tangent to the curve.
Example \(\PageIndex{8}\): Calculating a Surface Integral
Calculate surface integral
\[\iint_S 5 \, dS, \nonumber \]
where \(S\) is the surface with parameterization \(\vecs r(u,v) = \langle u, \, u^2, \, v \rangle\) for \(0 \leq u \leq 2\) and \(0 \leq v \leq u\).
Solution
Notice that this parameter domain \(D\) is a triangle, and therefore the parameter domain is not rectangular. This is not an issue though, because Equation \ref{scalar surface integrals} does not place any restrictions on the shape of the parameter domain.
To use Equation \ref{scalar surface integrals} to calculate the surface integral, we first find vectors \(\vecs t_u\) and \(\vecs t_v\). Note that \(\vecs t_u = \langle 1, 2u, 0 \rangle\) and \(\vecs t_v = \langle 0,0,1 \rangle\). Therefore,
\[\vecs t_u \times \vecs t_v = \begin{vmatrix} \mathbf{\hat i} & \mathbf{\hat j} & \mathbf{\hat k} \nonumber \\ 1 & 2u & 0 \nonumber \\ 0 & 0 & 1 \end{vmatrix} = \langle 2u, \, -1, \, 0 \rangle\ \nonumber \]
and
\[||\vecs t_u \times \vecs t_v|| = \sqrt{1 + 4u^2}. \nonumber \]
By Equation \ref{scalar surface integrals},
\[\begin{align*} \iint_S 5 \, dS &= 5 \iint_D \sqrt{1 + 4u^2} \, dA \\
&= 5 \int_0^2 \int_0^u \sqrt{1 + 4u^2} \, dv \, du = 5 \int_0^2 u \sqrt{1 + 4u^2}\, du \\
&= 5 \left[\dfrac{(1+4u^2)^{3/2}}{3} \right]_0^2 \\
&= \dfrac{5(17^{3/2}-1)}{3} \approx 115.15. \end{align*}\]
Example \(\PageIndex{9}\): Calculating the Surface Integral of a Cylinder
Calculate surface integral \[\iint_S (x + y^2) \, dS, \nonumber \] where \(S\) is cylinder \(x^2 + y^2 = 4, \, 0 \leq z \leq 3\) (Figure \(\PageIndex{15}\)).
Solution
To calculate the surface integral, we first need a parameterization of the cylinder.
The tangent vectors are \(\vecs t_u = \langle \sin u, \, \cos u, \, 0 \rangle\) and \(\vecs t_v = \langle 0,0,1 \rangle\). Then,
\[\vecs t_u \times \vecs t_v = \begin{vmatrix} \mathbf{\hat i} & \mathbf{\hat j} & \mathbf{\hat k} \\ -\sin u & \cos u & 0 \\ 0 & 0 & 1 \end{vmatrix} = \langle \cos u, \, \sin u, \, 0 \rangle \nonumber \]
and \(||\vecs t_u \times \vecs t_v || = \sqrt{\cos^2 u + \sin^2 u} = 1\).
- Hint
-
Use Equation \ref{scalar surface integrals}.
- Answer
-
24
Example \(\PageIndex{10}\): Calculating the Surface Integral of a Piece of a Sphere
Calculate surface integral \[\iint_S f(x,y,z)\,dS, \nonumber \] where \(f(x,y,z) = z^2\) and \(S\) is the surface that consists of the piece of sphere \(x^2 + y^2 + z^2 = 4\) that lies on or above plane \(z = 1\) and the disk that is enclosed by intersection plane \(z = 1\) and the given sphere (Figure \(\PageIndex{16}\)).
Solution
Notice that \(S\) is not smooth but is piecewise smooth; \(S\) can be written as the union of its base \(S_1\) and its spherical top \(S_2\), and both \(S_1\) and \(S_2\) are smooth. Therefore, to calculate
\[\iint_S z^2 dS, \nonumber \]
we write this integral as
\[\iint_{S_1} z^2 \,dS + \iint_{S_2} z^2 \,dS \nonumber \]
and we calculate integrals
\[\iint_{S_1} z^2 \,dS \nonumber \]
and
\[\iint_{S_2}Z^2 \,dS. \nonumber \]
First, we calculate \(\displaystyle \iint_{S_1} z^2 \,dS.\) To calculate this integral we need a parameterization of \(S_1\). This surface is a disk in plane \(z = 1\) centered at \((0,0,1)\). To parameterize this disk, we need to know its radius. Since the disk is formed where plane \(z = 1\) intersects sphere \(x^2 + y^2 + z^2 = 4\), we can substitute \(z = 1\) into equation \(x^2 + y^2 + z^2 = 4\):
\[x^2 + y^2 + 1 = 4 \Rightarrow x^2 + y^2 = 3. \nonumber \]
Therefore, the radius of the disk is \(\sqrt{3}\) and a parameterization of \(S_1\) is \(\vecs r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, 1 \rangle, \, 0 \leq u \leq \sqrt{3}, \, 0 \leq v \leq 2\pi\). The tangent vectors are \(\vecs t_u = \langle \cos v, \, \sin v, \, 0 \rangle \) and \(\vecs t_v = \langle -u \, \sin v, \, u \, \cos v, \, 0 \rangle\), and thus
\[\vecs t_u \times \vecs t_v = \begin{vmatrix} \mathbf{\hat i} & \mathbf{\hat j} & \mathbf{\hat k} \\ \cos v & \sin v & 0 \\ -u\sin v & u\cos v& 0 \end{vmatrix} = \langle 0, \, 0, u \, \cos^2 v + u \, \sin^2 v \rangle = \langle 0, 0, u \rangle. \nonumber \]
The magnitude of this vector is \(u\). Therefore,
\[\begin{align*} \iint_{S_1} z^2 \,dS &= \int_0^{\sqrt{3}} \int_0^{2\pi} f(r(u,v))||t_u \times t_v|| \, dv \, du \\
&= \int_0^{\sqrt{3}} \int_0^{2\pi} u \, dv \, du \\
&= 2\pi \int_0^{\sqrt{3}} u \, du \\
&= 2\pi \sqrt{3}. \end{align*}\]
Now we calculate
\[\iint_{S_2} \,dS. \nonumber \]
To calculate this integral, we need a parameterization of \(S_2\). The parameterization of full sphere \(x^2 + y^2 + z^2 = 4\) is
\[\vecs r(\phi, \theta) = \langle 2 \, \cos \theta \, \sin \phi, \, 2 \, \sin \theta \, \sin \phi, \, 2 \, \cos \phi \rangle, \, 0 \leq \theta \leq 2\pi, 0 \leq \phi \leq \pi. \nonumber \]
Since we are only taking the piece of the sphere on or above plane \(z = 1\), we have to restrict the domain of \(\phi\). To see how far this angle sweeps, notice that the angle can be located in a right triangle, as shown in Figure \(\PageIndex{17}\) (the \(\sqrt{3}\) comes from the fact that the base of \(S\) is a disk with radius \(\sqrt{3}\)). Therefore, the tangent of \(\phi\) is \(\sqrt{3}\), which implies that \(\phi\) is \(\pi / 6\). We now have a parameterization of \(S_2\):
\(\vecs r(\phi, \theta) = \langle 2 \, \cos \theta \, \sin \phi, \, 2 \, \sin \theta \, \sin \phi, \, 2 \, \cos \phi \rangle, \, 0 \leq \theta \leq 2\pi, \, 0 \leq \phi \leq \pi / 3.\)
The tangent vectors are \(\vecs t_{\phi} = \langle 2 \, \cos \theta \, \cos \phi, \, 2 \, \sin \theta \,\cos \phi, \, -2 \, \sin \phi \rangle\) and \(\vecs t_{\theta} = \langle - 2 \sin \theta \sin \phi, \, u\cos \theta \sin \phi, \, 0 \rangle\), and thus
\[\begin{align*} \vecs t_{\phi} \times \vecs t_{\theta} &= \begin{vmatrix} \mathbf{\hat i} & \mathbf{\hat j} & \mathbf{\hat k} \nonumber \\ 2 \cos \theta \cos \phi & 2 \sin \theta \cos \phi & -2\sin \phi \\ -2\sin \theta\sin\phi & 2\cos \theta \sin\phi & 0 \end{vmatrix} \\[4 pt]
&= \langle 4 \, \cos \theta \, \sin^2 \phi, \, 4 \, \sin \theta \, \sin^2 \phi, \, 4 \, \cos^2 \theta \, \cos \phi \, \sin \phi + 4 \, \sin^2 \theta \, \cos \phi \, \sin \phi \rangle \\[4 pt]
&= \langle 4 \, \cos \theta \, \sin^2 \phi, \, 4 \, \sin \theta \, \sin^2 \phi, \, 4 \, \cos \phi \, \sin \phi \rangle. \end{align*}\]
The magnitude of this vector is
\[\begin{align*} \vecs t_{\phi} \times \vecs t_{\theta} &= \sqrt{16 \, \cos^2\theta \, \sin^4\phi + 16 \, \sin^2\theta \, \sin^4 \phi + 16 \, \cos^2\phi \, \sin^2\phi} \\[4 pt]
&= 4 \sqrt{\sin^4\phi + \cos^2\phi \, \sin^2\phi}. \end{align*}\]
Therefore,
\[\begin{align*} \iint_{S_2} z \, dS &= \int_0^{\pi/6} \int_0^{2\pi} f (\vecs r(\phi, \theta))||\vecs t_{\phi} \times \vecs t_{\theta}|| \, d\theta \, d\phi \\
&= \int_0^{\pi/6} \int_0^{2\pi} 16 \, \cos^2\phi \sqrt{\sin^4\phi + \cos^2\phi \, \sin^2\phi} \, d\theta \, d\phi \\
&= 32 \pi \int_0^{\pi/6} \cos^2\phi \sqrt{\sin^4\phi + \cos^2\phi \, \sin^2 \phi} \, d\phi \\
&= 32 \pi \int_0^{\pi/6} \cos^2\phi \, \sin \phi \sqrt{\sin^2\phi + \cos^2\phi} \, d\phi \\
&= 32 \pi \int_0^{\pi/6} \cos^2\phi \, \sin \phi \, d\phi \\
&= 32\pi \left[- \dfrac{\cos^3 \phi}{3} \right]_0^{\pi/6} \\
&= 32 \pi \left[ \dfrac{1}{3} - \dfrac{\sqrt{3}}{8} \right] = \dfrac{32\pi}{3} - 4\sqrt{3}. \end{align*}\]
Since
\[\iint_S z^2 \,dS = \iint_{S_1}z^2 \,dS + \iint_{S_2}z^2 \,dS, \nonumber \]
we have
\[\iint_S z^2 \,dS = (2\pi - 4) \sqrt{3} + \dfrac{32\pi}{3}. \nonumber \]
Analysis
In this example we broke a surface integral over a piecewise surface into the addition of surface integrals over smooth subsurfaces. There were only two smooth subsurfaces in this example, but this technique extends to finitely many smooth subsurfaces.
Exercise \(\PageIndex{9}\)
Calculate line integral \(\displaystyle \iint_S (x - y) \, dS,\) where \(S\) is cylinder \(x^2 + y^2 = 1, \, 0 \leq z \leq 2\), including the circular top and bottom.
- Hint
-
Break the integral into three separate surface integrals.
- Answer
-
0
Scalar surface integrals have several real-world applications. Recall that scalar line integrals can be used to compute the mass of a wire given its density function. In a similar fashion, we can use scalar surface integrals to compute the mass of a sheet given its density function. If a thin sheet of metal has the shape of surface \(S\) and the density of the sheet at point \((x,y,z)\) is \(\rho(x,y,z)\) then mass \(m\) of the sheet is
\[\displaystyle m = \iint_S \rho (x,y,z) \,dS. \label{mass} \]
Example \(\PageIndex{11}\): Calculating the Mass of a Sheet
A flat sheet of metal has the shape of surface \(z = 1 + x + 2y\) that lies above rectangle \(0 \leq x \leq 4\) and \(0 \leq y \leq 2\). If the density of the sheet is given by \(\rho (x,y,z) = x^2 yz\), what is the mass of the sheet?
Solution
Let \(S\) be the surface that describes the sheet. Then, the mass of the sheet is given by \(\displaystyle m = \iint_S x^2 yx \, dS.\) To compute this surface integral, we first need a parameterization of \(S\). Since \(S\) is given by the function \(f(x,y) = 1 + x + 2y\), a parameterization of \(S\) is \(\vecs r(x,y) = \langle x, \, y, \, 1 + x + 2y \rangle, \, 0 \leq x \leq 4, \, 0 \leq y \leq 2\).
The tangent vectors are \(\vecs t_x = \langle 1,0,1 \rangle\) and \(\vecs t_y = \langle 1,0,2 \rangle\). Therefore, \(\vecs t_x + \vecs t_y = \langle -1,-2,1 \rangle\) and \(||\vecs t_x \times \vecs t_y|| = \sqrt{6}\).
By the definition of the line integral (Section 16.2), \[\begin{align*} m &= \iint_S x^2 yz \, dS \\[4pt]
&= \sqrt{6} \int_0^4 \int_0^2 x^2 y (1 + x + 2y) \, dy \,dx \\[4pt]
&= \sqrt{6} \int_0^4 \dfrac{22x^2}{3} + 2x^3 \,dx \\[4pt]
&= \dfrac{2560 \sqrt{6}}{9} \approx 696.74. \end{align*}\]
Exercise \(\PageIndex{10}\)
A piece of metal has a shape that is modeled by paraboloid \(z = x^2 + y^2, \, 0 \leq z \leq 4,\) and the density of the metal is given by \(\rho (x,y,z) = z + 1\). Find the mass of the piece of metal.
- Hint
-
The mass of a sheet is given by Equation \ref{mass}. A useful parameterization of a paraboloid was given in a previous example.
- Answer
-
\(38.401 \pi \approx 120.640\)
Orientation of a Surface
Recall that when we defined a scalar line integral, we did not need to worry about an orientation of the curve of integration. The same was true for scalar surface integrals: we did not need to worry about an “orientation” of the surface of integration.
On the other hand, when we defined vector line integrals, the curve of integration needed an orientation. That is, we needed the notion of an oriented curve to define a vector line integral without ambiguity. Similarly, when we define a surface integral of a vector field, we need the notion of an oriented surface. An oriented surface is given an “upward” or “downward” orientation or, in the case of surfaces such as a sphere or cylinder, an “outward” or “inward” orientation.
Let S be a smooth surface. For any point \((x,y,z)\) on \(S\), we can identify two unit normal vectors \(\vecs N\) and \(-\vecs N\). If it is possible to choose a unit normal vector \(\vecs N\) at every point \((x,y,z)\) on \(S\) so that \(\vecs N\) varies continuously over \(S\), then \(S\) is “ orientable .” Such a choice of unit normal vector at each point gives the orientation of a surface \(S\). If you think of the normal field as describing water flow, then the side of the surface that water flows toward is the “negative” side and the side of the surface at which the water flows away is the “positive” side.
Closed surfaces such as spheres are orientable: if we choose the outward normal vector at each point on the surface of the sphere, then the unit normal vectors vary continuously. This is called the positive orientation of the closed surface (Figure \(\PageIndex{18}\)). We also could choose the inward normal vector at each point to give an “inward” orientation, which is the negative orientation of the surface.
A portion of the graph of any smooth function \(z = f(x,y)\) is also orientable. If we choose the unit normal vector that points “above” the surface at each point, then the unit normal vectors vary continuously over the surface. We could also choose the unit normal vector that points “below” the surface at each point. To get such an orientation, we parameterize the graph of \(f\) in the standard way: \(\vecs r(x,y) = \langle x,\, y, \, f(x,y)\rangle\), where \(x\) and \(y\) vary over the domain of \(f\). Then, \(\vecs t_x = \langle 1,0,f_x \rangle\) and \(\vecs t_y = \langle 0,1,f_y \rangle \), and therefore the cross product \(\vecs t_x \times \vecs t_y\) (which is normal to the surface at any point on the surface) is \(\langle -f_x, \, -f_y, \, 1 \rangle \)Since the \(z\)-component of this vector is one, the corresponding unit normal vector points “upward,” and the upward side of the surface is chosen to be the “positive” side.
Let \(S\) be a smooth orientable surface with parameterization \(\vecs r(u,v)\). For each point \(\vecs r(a,b)\) on the surface, vectors \(\vecs t_u\) and \(\vecs t_v\) lie in the tangent plane at that point. Vector \(\vecs t_u \times \vecs t_v\) is normal to the tangent plane at \(\vecs r(a,b)\) and is therefore normal to \(S\) at that point. Therefore, the choice of unit normal vector
\[\vecs N = \dfrac{\vecs t_u \times \vecs t_v}{||\vecs t_u \times \vecs t_v||} \nonumber \]
gives an orientation of surface \(S\).
Example \(\PageIndex{12}\): Choosing an Orientation
Give an orientation of cylinder \(x^2 + y^2 = r^2, \, 0 \leq z \leq h\).

The tangent vectors are \(\vecs t_u = \langle -r \, \sin u, \, r \, \cos u, \, 0 \rangle \) and \(\vecs t_v = \langle 0,0,1 \rangle\). To get an orientation of the surface, we compute the unit normal vector
\[\vecs N = \dfrac{\vecs t_u \times \vecs t_v}{||\vecs t_u \times \vecs t_v||} \nonumber \]
In this case, \(\vecs t_u \times \vecs t_v = \langle r \, \cos u, \, r \, \sin u, \, 0 \rangle\) and therefore
\[||\vecs t_u \times \vecs t_v|| = \sqrt{r^2 \cos^2 u + r^2 \sin^2 u} = r. \nonumber \]
An orientation of the cylinder is
\[\vecs N(u,v) = \dfrac{\langle r \, \cos u, \, r \, \sin u, \, 0 \rangle }{r} = \langle \cos u, \, \sin u, \, 0 \rangle. \nonumber \]
Notice that all vectors are parallel to the \(xy\)-plane, which should be the case with vectors that are normal to the cylinder. Furthermore, all the vectors point outward, and therefore this is an outward orientation of the cylinder (Figure \(\PageIndex{19}\)).
Exercise \(\PageIndex{11}\)
Give the “upward” orientation of the graph of \(f(x,y) = xy\).
- Hint
-
Parameterize the surface and use the fact that the surface is the graph of a function.
- Answer
-
\[\vecs{N}(x,y) = \left\langle \dfrac{-y}{\sqrt{1+x^2+y^2}}, \, \dfrac{-x}{\sqrt{1+x^2+y^2}}, \, \dfrac{1}{\sqrt{1+x^2+y^2}} \right\rangle \nonumber \]
Since every curve has a “forward” and “backward” direction (or, in the case of a closed curve, a clockwise and counterclockwise direction), it is possible to give an orientation to any curve. Hence, it is possible to think of every curve as an oriented curve. This is not the case with surfaces, however. Some surfaces cannot be oriented; such surfaces are called nonorientable .  Some surfaces are twisted in such a fashion that there is no well-defined notion of an “inner” or “outer” side.
The classic example of a nonorientable surface is the Möbius strip. To create a Möbius strip, take a rectangular strip of paper, give the piece of paper a half-twist, and the glue the ends together (Figure \(\PageIndex{20}\)). Because of the half-twist in the strip, the surface has no “outer” side or “inner” side. If you imagine placing a normal vector at a point on the strip and having the vector travel all the way around the band, then (because of the half-twist) the vector points in the opposite direction when it gets back to its original position. Therefore, the strip really only has one side.
Since some surfaces are nonorientable , it is not possible to define a vector surface integral on all piecewise smooth surfaces. This is in contrast to vector line integrals, which can be defined on any piecewise smooth curve.
Surface Integral of a Vector Field
With the idea of orientable surfaces in place, we are now ready to define a surface integral of a vector field . The definition is analogous to the definition of the flux of a vector field along a plane curve. Recall that if \(\vecs{F}\) is a two-dimensional vector field and \(C\) is a plane curve, then the definition of the flux of \(\vecs{F}\) along \(C\) involved chopping \(C\) into small pieces, choosing a point inside each piece, and calculating \(\vecs{F} \cdot \vecs{N}\) at the point (where \(\vecs{N}\) is the unit normal vector at the point). The definition of a surface integral of a vector field proceeds in the same fashion, except now we chop surface \(S\) into small pieces, choose a point in the small (two-dimensional) piece, and calculate \(\vecs{F} \cdot \vecs{N}\) at the point.
To place this definition in a real-world setting, let \(S\) be an oriented surface with unit normal vector \(\vecs{N}\). Let \(\vecs{v}\) be a velocity field of a fluid flowing through \(S\), and suppose the fluid has density \(\rho(x,y,z)\) Imagine the fluid flows through \(S\), but \(S\) is completely permeable so that it does not impede the fluid flow (Figure \(\PageIndex{21}\)). The mass flux of the fluid is the rate of mass flow per unit area. The mass flux is measured in mass per unit time per unit area. How could we calculate the mass flux of the fluid across \(S\)?
The rate of flow, measured in mass per unit time per unit area, is \(\rho \vecs N\). To calculate the mass flux across \(S\), chop \(S\) into small pieces \(S_{ij}\). If \(S_{ij}\) is small enough, then it can be approximated by a tangent plane at some point \(P\) in \(S_{ij}\). Therefore, the unit normal vector at \(P\) can be used to approximate \(\vecs N(x,y,z)\) across the entire piece \(S_{ij}\) because the normal vector to a plane does not change as we move across the plane. The component of the vector \(\rho v\) at P in the direction of \(\vecs{N}\) is \(\rho \vecs v \cdot \vecs N\) at \(P\). Since \(S_{ij}\) is small, the dot product \(\rho v \cdot N\) changes very little as we vary across \(S_{ij}\) and therefore \(\rho \vecs v \cdot \vecs N\) can be taken as approximately constant across \(S_{ij}\). To approximate the mass of fluid per unit time flowing across \(S_{ij}\) (and not just locally at point \(P\)), we need to multiply \((\rho \vecs v \cdot \vecs N) (P)\) by the area of \(S_{ij}\). Therefore, the mass of fluid per unit time flowing across \(S_{ij}\) in the direction of \(\vecs{N}\) can be approximated by \((\rho \vecs v \cdot \vecs N)\Delta S_{ij}\) where \(\vecs{N}\), \(\rho\) and \(\vecs{v}\) are all evaluated at \(P\) (Figure \(\PageIndex{22}\)). This is analogous to the flux of two-dimensional vector field \(\vecs{F}\) across plane curve \(C\), in which we approximated flux across a small piece of \(C\) with the expression \((\vecs{F} \cdot \vecs{N}) \,\Delta s\). To approximate the mass flux across \(S\), form the sum
\[\sum_{i=1}m \sum_{j=1}^n (\rho \vecs{v} \cdot \vecs{N}) \Delta S_{ij}. \nonumber \]
As pieces \(S_{ij}\) get smaller, the sum
\[\sum_{i=1}m \sum_{j=1}^n (\rho \vecs{v} \cdot \vecs{N}) \Delta S_{ij} \nonumber \]
gets arbitrarily close to the mass flux. Therefore, the mass flux is
\[\iint_s \rho \vecs v \cdot \vecs N \, dS = \lim_{m,n\rightarrow\infty} \sum_{i=1}^m \sum_{j=1}^n (\rho \vecs{v} \cdot \vecs{N}) \Delta S_{ij}. \nonumber \]
This is a surface integral of a vector field. Letting the vector field \(\rho \vecs{v}\) be an arbitrary vector field \(\vecs{F}\) leads to the following definition.
Definition: Surface Integrals
Let \(\vecs{F}\) be a continuous vector field with a domain that contains oriented surface \(S\) with unit normal vector \(\vecs{N}\). The surface integral of \(\vecs{F}\) over \(S\) is
\[\iint_S \vecs{F} \cdot d\vecs{S} = \iint_S \vecs{F} \cdot \vecs{N} \,dS. \label{surfaceI} \]
Notice the parallel between this definition and the definition of vector line integral \(\displaystyle \int_C \vecs F \cdot \vecs N\, dS\). A surface integral of a vector field is defined in a similar way to a flux line integral across a curve, except the domain of integration is a surface (a two-dimensional object) rather than a curve (a one-dimensional object). Integral \(\displaystyle \iint_S \vecs F \cdot \vecs N\, dS\) is called the flux of \(\vecs{F}\) across \(S\), just as integral \(\displaystyle \int_C \vecs F \cdot \vecs N\,dS\) is the flux of \(\vecs F\) across curve \(C\). A surface integral over a vector field is also called a flux integral .
Just as with vector line integrals, surface integral \(\displaystyle \iint_S \vecs F \cdot \vecs N\, dS\) is easier to compute after surface \(S\) has been parameterized. Let \(\vecs r(u,v)\) be a parameterization of \(S\) with parameter domain \(D\). Then, the unit normal vector is given by \(\vecs N = \dfrac{\vecs t_u \times \vecs t_v}{||\vecs t_u \times \vecs t_v||}\) and, from Equation \ref{surfaceI}, we have
\[\begin{align*} \int_C \vecs F \cdot \vecs N\, dS &= \iint_S \vecs F \cdot \dfrac{\vecs t_u \times \vecs t_v}{||\vecs t_u \times \vecs t_v||} \,dS \\[4pt]
&= \iint_D \left(\vecs F (\vecs r (u,v)) \cdot \dfrac{\vecs t_u \times \vecs t_v}{||\vecs t_u \times \vecs t_v||} \right) || \vecs t_u \times \vecs t_v || \,dA \\[4pt]
&= \iint_D (\vecs F(\vecs r(u,v)) \cdot (\vecs t_u \times \vecs t_v))\,dA. \end{align*}\]
Therefore, to compute a surface integral over a vector field we can use the equation
\[\iint_S \vecs F \cdot \vecs N\, dS = \iint_D (\vecs F (\vecs r (u,v)) \cdot (\vecs t_u \times \vecs t_v)) \,dA. \label{equation 5} \]
Example \(\PageIndex{13}\): Calculating a Surface Integral
Calculate the surface integral
\[\iint_S \vecs F \cdot \vecs N\,dS, \nonumber \]
where \(\vecs{F} = \langle -y,x,0\rangle\) and \(S\) is the surface with parameterization
\[\vecs r(u,v) = \langle u,v^2 - u, \, u + v\rangle, \, 0 \leq u \leq 3, \, 0 \leq v \leq 4. \nonumber \]
Solution
The tangent vectors are \(\vecs t_u = \langle 1,-1,1\rangle\) and \(\vecs t_v = \langle 0,2v,1\rangle\). Therefore,
\[\vecs t_u \times \vecs t_v = \langle -1 -2v, -1, 2v\rangle. \nonumber \]
By Equation \ref{equation 5},
\[ \begin{align*} \iint_S \vecs F \cdot d\vecs S &= \int_0^4 \int_0^3 F (\vecs r(u,v)) \cdot (\vecs t_u \times \vecs t_v) \, du \,dv \\[4pt] &= \int_0^4 \int_0^3 \langle u - v^2, \, u, \, 0\rangle \cdot \langle -1 -2v, \, -1, \, 2v\rangle \, du\,dv \\[4pt] &= \int_0^4 \int_0^3 [(u - v^2)(-1-2v) - u] \, du\,dv \\[4pt] &= \int_0^4 \int_0^3 (2v^3 + v^2 - 2uv - 2u) \, du\,dv \\[4pt] &= \int_0^4 \left.[2v^3u + v^2u - vu^2 - u^2]\right|_0^3 \, dv \\[4pt] &= \int_0^4 (6v^3 + 3v^2 - 9v - 9) \, dv \\[4pt] &= \left[ \dfrac{3v^4}{2} + v^3 - \dfrac{9v^2}{2} - 9v\right]_0^4\\[4pt] &= 340. \end{align*}\]
Therefore, the flux of \(\vecs{F}\) across \(S\) is 340.
Exercise \(\PageIndex{12}\)
Calculate surface integral \[\iint_S \vecs F \cdot \vecs N \, dS, \nonumber \] where \(\vecs F = \langle 0, -z, y \rangle\) and \(S\) is the portion of the unit sphere in the first octant with outward orientation.
- Hint
-
Use Equation \ref{equation 5}.
- Answer
-
0
Example \(\PageIndex{14}\): Calculating Mass Flow Rate
Let \(\vecs v(x,y,z) = \langle 2x, \, 2y, \, z\rangle\) represent a velocity field (with units of meters per second) of a fluid with constant density 80 kg/m 3 . Let \(S\) be hemisphere \(x^2 + y^2 + z^2 = 9\) with \(z \leq 0\) such that \(S\) is oriented outward. Find the mass flow rate of the fluid across \(S\).
Solution
A parameterization of the surface is
\[\vecs r(\phi, \theta) = \langle 3 \, \cos \theta \, \sin \phi, \, 3 \, \sin \theta \, \sin \phi, \, 3 \, \cos \phi \rangle, \, 0 \leq \theta \leq 2\pi, \, 0 \leq \phi \leq \pi/2. \nonumber \]
As in Example \(\PageIndex{7}\), the tangent vectors are \(\vecs t_{\theta} = \langle -3 \, \sin \theta \, \sin \phi, \, 3 \, \cos \theta \, \sin \phi, \, 0 \rangle \) and \( \vecs t_{\phi} = \langle 3 \, \cos \theta \, \cos \phi, \, 3 \, \sin \theta \, \cos \phi, \, -3 \, \sin \phi \rangle,\) and their cross product is
\[\vecs t_{\phi} \times \vecs t_{\theta} = \langle 9 \, \cos \theta \, \sin^2 \phi, \, 9 \, \sin \theta \, \sin^2 \phi, \, 9 \, \sin \phi \, \cos \phi \rangle. \nonumber \]
Notice that each component of the cross product is positive, and therefore this vector gives the outward orientation. Therefore we use the orientation
\(\vecs N = \langle 9 \, \cos \theta \, \sin^2 \phi, \, 9 \, \sin \theta \, \sin^2 \phi, \, 9 \, \sin \phi \, \cos \phi \rangle \)
for the sphere.
By Equation \ref{surfaceI},
\[\begin{align*} \iint_S \rho v \cdot \,dS &= 80 \int_0^{2\pi} \int_0^{\pi/2} v (r(\phi, \theta)) \cdot (t_{\phi} \times t_{\theta}) \, d\phi \, d\theta \\
&= 80 \int_0^{2\pi} \int_0^{\pi/2} \langle 6 \, \cos \theta \, \sin \phi, \, 6 \, \sin \theta \, \sin \phi, \, 3 \, \cos \phi \rangle \cdot \langle 9 \, \cos \theta \, \sin^2 \phi, \, 9 \, \sin \theta \, \sin^2 \phi, \, 9 \, \sin \phi \, \cos \phi \rangle \, d\phi \, d\theta \\
&= 80 \int_0^{2\pi} \int_0^{\pi/2} 54 \, \sin^3 \phi + 27 \, \cos^2 \phi \, \sin \phi \, d\phi \, d\theta \\
&= 80 \int_0^{2\pi} \int_0^{\pi/2} 54 (1 - \cos^2\phi) \, \sin \phi + 27 \cos^2\phi \, \sin \phi \, d\phi \, d\theta \\
&= 80 \int_0^{2\pi} \int_0^{\pi/2} 54\, \sin \phi - 27 \, \cos^2 \phi \, \sin \phi \, d\phi \,d\theta \\
&= 80 \int_0^{2\pi} \Big[-54 \, \cos \phi + 9 \, \cos^3 \phi \Big]_{\phi=0}^{\phi=2\pi} \, d\theta \\
&=80 \int_0^{2\pi} 45 \, d\theta \\
&= 7200\pi.\end{align*} \nonumber \]
Therefore, the mass flow rate is \(7200\pi \, \text{kg/sec/m}^2\).
Exercise \(\PageIndex{13}\)
Let \(\vecs v(x,y,z) = \langle x^2 + y^2, \, z, \, 4y \rangle\) m/sec represent a velocity field of a fluid with constant density 100 kg/m 3 .  Calculate the mass flux of the fluid across \(S\).
- Hint
-
Use Equation \ref{surfaceI}.
- Answer
-
400 kg/sec/m
In Example \(\PageIndex{14}\), we computed the mass flux, which is the rate of mass flow per unit area. If we want to find the flow rate (measured in volume per time) instead, we can use flux integral
\[\iint_S \vecs v \cdot \vecs N \, dS, \nonumber \]
which leaves out the density. Since the flow rate of a fluid is measured in volume per unit time, flow rate does not take mass into account. Therefore, we have the following characterization of the flow rate of a fluid with velocity \(\vecs v\) across a surface \(S\):
\[\text{Flow rate of fluid across S} = \iint_S \vecs v \cdot d\vecs S. \nonumber \]
To compute the flow rate of the fluid in Example \(\PageIndex{14}\), we simply remove the density constant, which gives a flow rate of \(90 \pi \, m^3/sec\).
Both mass flux and flow rate are important in physics and engineering. Mass flux measures how much mass is flowing across a surface; flow rate measures how much volume of fluid is flowing across a surface.
In addition to modeling fluid flow, surface integrals can be used to model heat flow. Suppose that the temperature at point \((x,y,z)\) in an object is \(T(x,y,z)\). Then the heat flow is a vector field proportional to the negative temperature gradient in the object. To be precise, the heat flow is defined as vector field \(F = - k \nabla T\), where the constant k is the thermal conductivity of the substance from which the object is made (this constant is determined experimentally). The rate of heat flow across surface S in the object is given by the flux integral
\[\iint_S \vecs F \cdot d\vecs S = \iint_S -k \vecs \nabla T \cdot d\vecs S. \nonumber \]
Example \(\PageIndex{15}\): Calculating Heat Flow
A cast-iron solid cylinder is given by inequalities \(x^2 + y^2 \leq 1, \, 1 \leq z \leq 4\). The temperature at point \((x,y,z)\) in a region containing the cylinder is \(T(x,y,z) = (x^2 + y^2)z\). Given that the thermal conductivity of cast iron is 55, find the heat flow across the boundary of the solid if this boundary is oriented outward.
Solution
Let \(S\) denote the boundary of the object. To find the heat flow, we need to calculate flux integral \[\iint_S -k\vecs \nabla T \cdot d\vecs S. \nonumber \] Notice that \(S\) is not a smooth surface but is piecewise smooth, since \(S\) is the union of three smooth surfaces (the circular top and bottom, and the cylindrical side). Therefore, we calculate three separate integrals, one for each smooth piece of \(S\). Before calculating any integrals, note that the gradient of the temperature is \(\vecs \nabla T = \langle 2xz, \, 2yz, \, x^2 + y^2 \rangle\).
First we consider the circular bottom of the object, which we denote \(S_1\). We can see that \(S_1\) is a circle of radius 1 centered at point \((0,0,1)\) sitting in plane \(z = 1\).
Therefore,
\(\vecs t_u = \langle -v \, \sin u, \, v \, \cos u, \, 0 \rangle\) and \(\vecs t_v = \langle \cos u, \, v \, \sin u, \, 0 \rangle\), and \(\vecs t_u \times \vecs t_v = \langle 0, \, 0, -v \, \sin^2 u - v \, \cos^2 u \rangle = \langle 0, \, 0, -v \rangle\).
Since the surface is oriented outward and \(S_1\) is the bottom of the object, it makes sense that this vector points downward. By Equation \ref{equation 5} the heat flow across \(S_1\) is
\[ \begin{align*}\iint_{S_1} -k \vecs \nabla T \cdot d\vecs S &= - 55 \int_0^{2\pi} \int_0^1 \vecs \nabla T(u,v) \cdot (\vecs t_u \times \vecs t_v) \, dv\, du \\[4pt] &= - 55 \int_0^{2\pi} \int_0^1 \langle 2v \, \cos u, \, 2v \, \sin u, \, v^2 \cos^2 u + v^2 \sin^2 u \rangle \cdot \langle 0,0, -v\rangle \, dv \,du \\[4pt] &= - 55 \int_0^{2\pi} \int_0^1 \langle 2v \, \cos u, \, 2v \, \sin u, \, v^2\rangle \cdot \langle 0, 0, -v \rangle \, dv\, du \\[4pt] &= - 55 \int_0^{2\pi} \int_0^1 -v^3 \, dv\, du \\[4pt] &= - 55 \int_0^{2\pi} -\dfrac{1}{4} du \\[4pt] &= \dfrac{55\pi}{2}.\end{align*}\]
Now let’s consider the circular top of the object, which we denote \(S_2\). We see that \(S_2\) is a circle of radius 1 centered at point \((0,0,4)\), sitting in plane \(z = 4\).
Therefore, \(\vecs t_u = \langle -v \, \sin u, \, v \, \cos u, \, 0 \rangle\) and \(\vecs t_v = \langle \cos u, \, v \, \sin u, \, 0 \rangle \), and \(\vecs t_u \times \vecs t_v = \langle 0, \, 0, -v \, \sin^2 u - v \, \cos^2 u \rangle = \langle 0,0,-v\rangle\).
Since the surface is oriented outward and \(S_1\) is the top of the object, we instead take vector \(\vecs t_v \times \vecs t_u = \langle 0,0,v\rangle\). By Equation \ref{equation 5}, the heat flow across \(S_1\) is
\[ \begin{align*}\iint_{S_2} -k \vecs \nabla T \cdot d\vecs S &= - 55 \int_0^{2\pi} \int_0^1 \vecs \nabla T(u,v) \cdot\, (\vecs t_u \times \vecs t_v) \, dv\, du \\[4pt]
&= - 55 \int_0^{2\pi} \int_0^1 \langle 8v \, \cos u, \, 8v \, \sin u, \, v^2 \cos^2 u + v^2 \sin^2 u \rangle \cdot \langle 0,0, -v\rangle \, dv\,du \\[4pt]
&= - 55 \int_0^{2\pi} \int_0^1 \langle 8v \, \cos u, \, 8v \, \sin u, \, v^2\rangle \cdot \langle 0, 0, -v \rangle\, \, dv \,du\\[4pt]
&= - 55 \int_0^{2\pi} \int_0^1 -v^3 \, dv \,du = - 55 \int_0^{2\pi} -\dfrac{1}{4} \,du = - \dfrac{55\pi}{2}.\end{align*}\]
Last, let’s consider the cylindrical side of the object.  By Example \(\PageIndex{9}\), we know that \(\vecs t_u \times \vecs t_v = \langle \cos u, \, \sin u, \, 0 \rangle\). By Equation \ref{equation 5},
\[ \begin{align*} \iint_{S_3} -k \vecs \nabla T \cdot d\vecs S &= - 55 \int_0^{2\pi} \int_1^4 \vecs \nabla T(u,v) \cdot (\vecs t_u \times \vecs t_v) \, dv\, du \\[4pt]
&= - 55 \int_0^{2\pi} \int_1^4 \langle 2v \, \cos u, \, 2v \, \sin u, \, \cos^2 u + \sin^2 u \rangle \cdot \langle \cos u, \, \sin u, \, 0 \rangle \, dv\, du \\[4pt]
&= - 55 \int_0^{2\pi} \int_0^1 \langle 2v \, \cos^2 u, \, 2v \, \sin u, \, 1 \rangle \cdot \langle \cos u, \, \sin u, \, 0 \rangle \, dv\,\, du \\[4pt]
&= - 55 \int_0^{2\pi} \int_0^1 (2v \, \cos^2 u + 2v \, \sin^2 u ) \, dv \,du \\[4pt]
&= - 55 \int_0^{2\pi} \int_0^1 2v \, dv \,du \\[4pt]
&= -55 \int_0^{2\pi} du \\[4pt]
&= -110\pi. \end{align*}\]
Therefore, the rate of heat flow across \(S\) is
\[\dfrac{55\pi}{2} - \dfrac{55\pi}{2} - 110\pi = -110\pi. \nonumber \]
Exercise \(\PageIndex{14}\)
A cast-iron solid ball is given by inequality \(x^2 + y^2 + z^2 \leq 1\). The temperature at a point in a region containing the ball is \(T(x,y,z) = \dfrac{1}{3}(x^2 + y^2 + z^2)\). Find the heat flow across the boundary of the solid if this boundary is oriented outward.
- Hint
-
Follow the steps of Example \(\PageIndex{15}\).
- Answer
-
\(-\dfrac{440\pi}{3}\)
Key Concepts
- Surfaces can be parameterized, just as curves can be parameterized. In general, surfaces must be parameterized with two parameters.
- Surfaces can sometimes be oriented, just as curves can be oriented. Some surfaces, such as a Möbius strip, cannot be oriented.
-
A surface integral is like a line integral in one higher dimension. The domain of integration of a surface integral is a surface in a plane or space, rather than a curve in a plane or space.
- The integrand of a surface integral can be a scalar function or a vector field. To calculate a surface integral with an integrand that is a function, use Equation \ref{scalar surface integrals}. To calculate a surface integral with an integrand that is a vector field, use Equation \ref{equation 5}.
- If \(S\) is a surface, then the area of \(S\) is \[\iint_S \, dS. \nonumber \]
Key Equations
- S calar surface integral
\[\iint_S f(x,y,z) \,dS = \iint_D f (\vecs r(u,v)) ||\vecs t_u \times \vecs t_v||\,dA \nonumber \]
- Flux integral
\[\iint_S \vecs F \cdot \vecs N \, dS = \iint_S \vecs F \cdot d\vecs S = \iint_D \vecs F (\vecs r (u,v)) \cdot (\vecs t_u \times \vecs t_v) \, dA \nonumber \]
Glossary
- flux integral
- another name for a surface integral of a vector field; the preferred term in physics and engineering
- grid curves
- curves on a surface that are parallel to grid lines in a coordinate plane
- heat flow
- a vector field proportional to the negative temperature gradient in an object
- mass flux
- the rate of mass flow of a fluid per unit area, measured in mass per unit time per unit area
- orientation of a surface
- if a surface has an “inner” side and an “outer” side, then an orientation is a choice of the inner or the outer side; the surface could also have “upward” and “downward” orientations
- parameter domain (parameter space)
- the region of the \(uv\)-plane over which the parameters \(u\) and \(v\) vary for parameterization \(\vecs r(u,v) = \langle x(u,v), \, y(u,v), \, z(u,v)\rangle\)
- parameterized surface (parametric surface)
- a surface given by a description of the form \(\vecs r(u,v) = \langle x(u,v), \, y(u,v), \, z(u,v)\rangle\), where the parameters \(u\) and \(v\) vary over a parameter domain in the \(uv\)-plane
- regular parameterization
- parameterization \(\vecs r(u,v) = \langle x(u,v), \, y(u,v), \, z(u,v)\rangle\) such that \(r_u \times r_v\) is not zero for point \((u,v)\) in the parameter domain
- surface area
- the area of surface \(S\) given by the surface integral \[\iint_S \,dS \nonumber \]
- surface integral
- an integral of a function over a surface
- surface integral of a scalar-valued function
- a surface integral in which the integrand is a scalar function
- surface integral of a vector field
- a surface integral in which the integrand is a vector field
It follows from Example \(\PageIndex{1}\) that we can parameterize all cylinders of the form \(x^2 + y^2 = R^2\).
We can also find different types of surfaces given their parameterization, or we can find a parameterization when we are given a surface.

Solution
Notice that if \(u\) is held constant, then the resulting curve is a circle of radius \(u\) in plane \(z = u\). Therefore, as \(u\) increases, the radius of the resulting circle increases. If \(v\) is held constant, then the resulting curve is a vertical parabola. Therefore, we expect the surface to be an elliptic paraboloid. To confirm this, notice that
\[\begin{align*} x^2 + y^2 &= (u \, \cos v)^2 + (u \, \sin v)^2 \\[4pt] &= u^2 \cos^2 v + u^2 sin^2 v \\[4pt] &= u^2 \\[4pt] &=z\end{align*}\]
Therefore, the surface is the elliptic paraboloid \(x^2 + y^2 = z\) (Figure \(\PageIndex{3}\)).

Hold \(u\) constant and see what kind of curves result. Imagine what happens as \(u\) increases or decreases.
Cone \(x^2 + y^2 = z^2\)


7.E: Electronic Structure and Periodic Properties- Homework
Turn in your answers for the following questions - show your work
-
Write the electron configuration (ie: 1s
2
, 2s
1
), orbital energy diagram, and orbital box diagram for,
- H, and Na
- Be and Ca
- C and Ge
- Cl 1 - and K 1+
The Following Questions are for your practice - Do Not Turn In. They include answers so you can check your work
Electromagnetic Energy
Q6.1.2
An FM radio station found at 103.1 on the FM dial broadcasts at a frequency of 1.031 × 10 8 s −1 (103.1 MHz). What is the wavelength of these radio waves in meters?
S6.1.2
\(λ=\dfrac{c}{ν}\)
\(λ = \dfrac{2.998 \times10^{8}\: \dfrac{m}{s}}{1.031 \times10^{8}\: \dfrac{1}{s}} = 2.908\:m\)
Q6.1.4
A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be released by an electron in a mercury atom to produce a photon of this light?
S6.1.4
\(E= \dfrac{hc}{λ}\)
\(E= \dfrac{(2.998 \times 10^{8} \dfrac{m}{s})\: (6.6262 \times 10^{-34} Js)}{4.358 \times 10^{-7} m}\)
\(=4.56\times 10^{-19}J\)
The Bohr Model
Q6.2.11
Using the Bohr model, determine the energy in joules of the photon produced when an electron in a Li2+ ion moves from the orbit with n = 2 to the orbit with n = 1.
S6.2.11
1.471 × 10−17 J
Development of Quantum Theory
Q6.3.3
Describe the properties of an electron associated with each of the following four quantum numbers: n , l , m l , and m s .
S6.3.3
n determines the general range for the value of energy and the probable distances that the electron can be from the nucleus. l determines the shape of the orbital. m l determines the orientation of the orbitals of the same l value with respect to one another. m s determines the spin of an electron.
Q6.3.5
Identify the subshell in which electrons with the following quantum numbers are found:
- n = 2, l = 1
- n = 4, l = 2
- n = 6, l = 0
S6.3.5
(a) 2 p ; (b) 4 d ; (c) 6 s
Q6.3.11
Consider the orbitals shown here in outline.
- What is the maximum number of electrons contained in an orbital of type (x)? Of type (y)?
- How many orbitals of type (x) are found in a shell with n = 2? How many of type (y)?
- Write a set of quantum numbers for an electron in an orbital of type (x) in a shell with n = 4. Of an orbital of type (y) in a shell with n = 2.
- What are the possible l and m l values for an orbital of type (x)? Of type (y)?
S6.3.11
(a) x. 2, y. 2 ;(b) x. 1, y. 3 (c) x. 4 0 0 \(\dfrac{1}{2}\), y. 2 1 0 \(\dfrac{1}{2}\); (d) x. l = 0, m l = 0, y. l = 1, m l = –1, 0, or + 1
Q6.3.15
Write a set of quantum numbers for each of the electrons with an n of 4 in a Se atom.
S6.3.15
| n | l | m l | s |
|---|---|---|---|
| 4 | 0 | 0 | \(+\dfrac{1}{2}\) |
| 4 | 0 | 0 | \(−\dfrac{1}{2}\) |
| 4 | 1 | −1 | \(+\dfrac{1}{2}\) |
| 4 | 1 | 0 | \(+\dfrac{1}{2}\) |
| 4 | 1 | +1 | \(+\dfrac{1}{2}\) |
| 4 | 1 | −1 | \(−\dfrac{1}{2}\) |
Electronic Structure of Atoms (Electron Configurations)
Q6.4.3
Using complete subshell notation (1 s 2 2 s 2 2 p 6 , and so forth), predict the electron configuration of each of the following atoms:
- C
- P
S6.4.3
a.) 1s 2 2s 2 2p 2
b.) 1s 2 2s 2 2p 6 3s 2 3p 3
Q6.4.8
Use an orbital box diagram to describe the electron configuration of the valence shell of each of the following atoms:
- N
S6.4.8
(a)
Q6.4.12
Which ion with a +1 charge has the electron configuration 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 ? Which ion with a –2 charge has this configuration?
S6.4.12
Rb + , Se 2 −
Periodic Variations in Element Properties
Q6.5.1
Based on their positions in the periodic table, predict which has the smallest atomic radius: Mg, Sr, Si, Cl, I.
S6.5.5
Rb < Li < N < F
Q6.5.11
Based on their positions in the periodic table, list the following ions in order of increasing radius: K + , Ca 2 + , Al 3 + , Si 4 + .
S6.5.11
Si 4 + < Al 3 + < Ca 2 + < K +


14.4: Information Processing Theory- Memory, Encoding, and Storage
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Memory
Memory is an information processing system that we often compare to a computer. Memory is the set of processes used to encode, store, and retrieve information over different periods of time.
Encoding involves the input of information into the memory system. Storage is the retention of the encoded information. Retrieval, or getting the information out of memory and back into awareness, is the third function.
Encoding (Input of Information to Memory)
We get information into our brains through a process called encoding , which is the input of information into the memory system. Once we receive sensory information from the environment, our brains label or code it. We organize the information with other similar information and connect new concepts to existing concepts. Encoding information occurs through both automatic processing and effortful processing. For example, if someone asks you what you ate for lunch today, more than likely you could recall this information quite easily. This is known as automatic processing , or the encoding of details like time, space, frequency, and the meaning of words. Automatic processing is usually done without any conscious awareness.
Recalling the last time you studied for a test is another example of automatic processing. But what about the actual test material you studied? It probably required a lot of work and attention on your part to encode that information; this is known as effortful processing . When you first learn new skills such as driving a car, you have to put forth effort and attention to encode information about how to start a car, how to brake, how to handle a turn, and so on. Once you know how to drive, you can encode additional information about this skill automatically.
Storage (Retaining Information in Memory)
Once the information has been encoded, we have to retain it. Our brains take the encoded information and place it in storage. Storage is the creation of a permanent record of information. In order for a memory to go into storage (i.e., long-term memory), it has to pass through three distinct stages: Sensory Memory, Short-Term Memory, and finally Long-Term Memory. These stages were first proposed by Richard Atkinson and Richard Shiffrin (1968). Their model of human memory, called Atkinson-Shiffrin (A-S), is based on the belief that we process memories in the same way that a computer processes information.
Sensory Memory (First Stage of Storage)
In the Atkinson-Shiffrin model, stimuli from the environment are processed first in sensory memory , storage of brief sensory events, such as sights, sounds, and tastes. It is very brief storage—up to a couple of seconds. We are constantly bombarded with sensory information. We cannot absorb all of it, or even most of it.  For example, what was your professor wearing the last class period? As long as the professor was dressed appropriately, it does not really matter what they were wearing. Sensory information about sights, sounds, smells, and even textures, which we do not view as valuable information, we discard. If we view something as valuable, the information will move into our short-term memory system.
One study of sensory memory researched the significance of valuable information on short-term memory storage. J. R. Stroop discovered a memory phenomenon in the 1930s: you will name a color more easily if it appears printed in that color, which is called the Stroop effect.
The Stroop Effect describes why it is difficult for us to name a color when the word and the color of the word are different. To test this out a person is instructed not to read the words below, but to say the color the word is printed in. For example, upon seeing the word “yellow” in green print, they should say “green,” not “yellow.” This experiment is fun, but it’s not as easy as it seems.
Short-Term Memory or Working Memory (Second Stage of Storage)
Short-term memory is a temporary storage system that processes incoming sensory memory; sometimes it is called working memory. Short-term memory takes information from sensory memory and sometimes connects that memory to something already in long-term memory. Short-term memory storage lasts about 20 seconds. Think of short-term memory as the information you have displayed on your computer screen—a document, a spreadsheet, or a web page. Information in short-term memory either goes to long-term memory (when you save it to your hard drive) or it is discarded (when you delete a document or close a web browser).
George Miller (1956), in his research on the capacity of memory, found that most people can retain about seven items in short-term memory. Some remember five, some nine, so he called the capacity of short-term memory the range of seven items plus or minus two.
To explore the capacity and duration of short-term memory, two people can try this activity. One person reads the strings of random numbers below out loud to the other, beginning each string by saying, “Ready?” and ending each by saying, “Recall.” Then the second person should try to write down the string of numbers from memory.
This can be used to determine the longest string of digits that you can store. For most people, this will be close to seven, Miller’s famous seven plus or minus two. Recall is somewhat better for random numbers than for random letters (Jacobs, 1887) and is also often slightly better for information we hear ( acoustic encoding , which is the encoding of sounds) rather than what we see ( visual encoding , which is the encoding of images and words in particular) (Anderson, 1969).
Long-Term Memory (Third and Final Stage of Storage)
Long-term memory is the continuous storage of information. Unlike short-term memory, the storage capacity of long-term memory has no limits. It encompasses all the things you can remember that happened more than just a few minutes ago to all of the things that you can remember that happened days, weeks, and years ago. In keeping with the computer analogy, the information in your long-term memory would be like the information you have saved on the hard drive. It isn’t there on your desktop (your short-term memory), but you can pull up this information when you want it, at least most of the time. Not all long-term memories are strong memories. Some memories can only be recalled through prompts. For example, you might easily recall a fact— “What is the capital of the United States?”—or a procedure—“How do you ride a bike?”—but you might struggle to recall the name of the restaurant you had dinner at when you were on vacation in France last summer. A prompt, such as that the restaurant was named after its owner, who spoke to you about your shared interest in soccer, may help you recall (retrieve) the name of the restaurant.
Retrieval (Finding Memories)
So you have worked hard to encode via effortful processing (a lot of work and attention on your part in order to encode that information) and store some important information for your upcoming final exam. How do you get that information back out of storage when you need it? The act of getting information out of memory storage and back into conscious awareness is known as retrieval . This would be similar to finding and opening a paper you had previously saved on your computer’s hard drive. Now it’s back on your desktop, and you can work with it again. Our ability to retrieve information from long-term memory is vital to our everyday functioning. You must be able to retrieve information from memory in order to do everything from knowing how to brush your hair and teeth, to driving to work, to knowing how to perform your job once you get there.
Long-Term Memory Retrieval (Storage System): Recall, Recognition, Relearning, and Forgetting
There are three ways you can retrieve information out of your long-term memory storage system: recall, recognition, and relearning. Recall is what we most often think about when we talk about memory retrieval: it means you can access information without cues. For example, you would use recall for an essay test. Recognition happens when you identify information that you have previously learned after encountering it again. It involves a process of comparison. When you take a multiple-choice test, you are relying on recognition to help you choose the correct answer. The third form of retrieval is relearning , and it’s just what it sounds like, it involves learning information that you previously learned. Whitney took Spanish in high school, but after high school she did not have the opportunity to speak Spanish. Whitney is now 31, and her company has offered her an opportunity to work in their Mexico City office. In order to prepare herself, she enrolls in a Spanish course at the local community college. She’s surprised at how quickly she’s able to pick up the language after not speaking it for 13 years; this is an example of relearning.
Forgetting (It Wasn’t Locked In)
As we just learned, your brain must do some work (effortful processing) to encode information and move it into short-term, and ultimately long-term memory. This has strong implications for a student, as it can impact their learning – if one doesn’t work to encode and store information, it will likely be forgotten. Research indicates that people forget 80 percent of what they learn only a day later. This statistic may not sound very encouraging, given all that you’re expected to learn and remember as a college student. Really, though, it points to the importance of a study strategy other than waiting until the night before a final exam to review a semester’s worth of readings and notes. When you learn something new, the goal is to “lock it in” sooner rather than later, and move it from short-term memory to long-term memory, where it can be accessed when you need it (like at the end of the semester for your final exam or maybe years from now). The next section will explore a variety of strategies that can be used to process information more deeply and help improve retrieval. 18
Memory Strategies 19
Knowing What to Know How can you decide what to study and what you need to know? The answer is to prioritize what you’re trying to learn and memorize, rather than trying to tackle all of it. Below are some strategies to help you do this:
- Think about concepts rather than facts: Most of the time instructors are concerned about you learning about the key concepts in a subject or course rather than specific facts.
- Take cues from your instructor: Pay attention to what your instructor writes on the board, mentions repeatedly in class, or includes in study guides and handouts, they are likely core concepts that you’ll want to focus on.
- Look for key terms: Textbooks will often put key terms in bold or italics.
- Use summaries: Read end of chapter summaries, or write your own, to check your understanding of the main elements of the reading.
Transferring Information from Short-Term Memory to Long-Term Memory In the previous discussion of how memory works, the importance of making intentional efforts to transfer information from short-term to long-term memory was noted. Below are some strategies to facilitate this process:
- Start reviewing new material immediately : Remember that people typically forget a significant amount of new information within 24 hours of learning it.
- Study frequently for shorter periods of time : If you want to improve the odds of recalling course material by the time of an exam or in future class, try reviewing it a little bit every day.
Strengthening your Memory How can you work to strengthen your overall memory? Some people have stronger memories than others but memorizing new information takes work for anyone. Below are some strategies that can aid memory:
- Rehearsal: One strategy is rehearsal, or the conscious repetition of information to be remembered (Craik & Watkins, 1973). Academic learning comes with time and practice, and at some point the skills become second nature.
- Incorporate visuals: Visual aids like note cards, concept maps, and highlighted text are ways of making information stand out. These aids make the information to be memorized seem more manageable and less daunting.
-
Create mnemonics:
Memory devices known as mnemonics can help you retain information while only needing to remember a unique phrase or letter pattern that stands out. They are especially useful when we want to recall larger bits of information such as steps, stages, phases, and parts of a system (Bellezza, 1981). There are different types of mnemonic devices:
- Acronym: An acronym is a word formed by the first letter of each of the words you want to remember. Such as HOMES for the Great Lakes (Huron, Ontario, Michigan, Erie, and Superior)
- Acrostic: In an acrostic, you make a phrase of all the first letters of the words. For example, if you need to remember the order of mathematical operations, recalling the sentence “Please Excuse My Dear Aunt Sally” will help you, because the order of mathematical operations is Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
- Jingles: Rhyming tunes that contain key words related to the concept, such as “i before e, except after c” are jingles.
- Visual: Using a visual to help you remember is also useful. Such as the knuckle mnemonic shown in the image below to help you remember the number of days in each month. Months with 31 days are represented by the protruding knuckles and shorter months fall in the spots between knuckles.
- Chunking: Another strategy is chunking, where you organize information into manageable bits or chunks, such as turning a phone number you remember into chunks.
- Connect new information to old information: It’s easier to remember new information if you can connect it to old information, a familiar frame of reference, or a personal experience.
- Get quality sleep: Although some people require more or less sleep than the recommended amount, most people should aim for six to eight hours every night.
Contributors and Attributions
18. Memory and Information Processing by Laura Lucas is licensed under CC BY 4.0
19. Memory and Information Processing by Laura Lucas is licensed under CC BY 4.0


22.4: Urinary Tract
By the end of the section, you will be able to:
- Identify the ureters, urinary bladder, and urethra, as well as their location, structure, histology, and function
- Compare and contrast male and female urethras
- Describe the micturition reflex
Like the Digestive System with the digestive tract, the Urinary System also has a urinary tract, which includes the ureters, urinary bladder , and urethra . This section will start with urine excretion, tracing the flow of urine from the renal pelvis to ureter, urinary bladder, and urethra. Urine is a fluid of variable composition that requires specialized structures to remove it from the body safely and efficiently. Blood is filtered, and the filtrate is transformed into urine at a relatively constant rate throughout the day. This processed liquid is stored until a convenient time for excretion. All structures involved in the transport and storage of the urine are large enough to be visible to the naked eye. This transport and storage system not only stores the waste, but it protects the tissues from damage, prevents infection by foreign organisms, and for the male, provides reproductive functions.
Ureters
The kidneys and ureters are completely retroperitoneal, and the bladder has a peritoneal covering only over the dome. As urine is formed, it drains into the calyces of the kidney, which merge to form the funnel-shaped renal pelvis in the hilum of each kidney. The hilum narrows to become the ureter of each kidney. As urine passes through the ureter, it does not passively drain into the bladder but rather is propelled by waves of peristalsis. As the ureters enter the pelvis, they sweep laterally, hugging the pelvic walls. As they approach the bladder, they turn medially and pierce the bladder wall obliquely. This is important because it creates an one-way valve (a physiological sphincter rather than an anatomical sphincter ) that allows urine into the bladder but prevents reflux of urine from the bladder back into the ureter. Children born lacking this oblique course of the ureter through the bladder wall are susceptible to “vesicoureteral reflux,” which dramatically increases their risk of serious UTI. Pregnancy also increases the likelihood of reflux and UTI.
The ureters are approximately 30 cm long (Figure 22.4.1). The inner mucosa is lined with transitional epithelium and scattered goblet cells that secrete protective mucus. The muscular layer of the ureter consists of longitudinal and circular smooth muscles that create the peristaltic contractions to move the urine into the bladder without the aid of gravity.
Urinary Bladder
The urinary bladder collects urine from both ureters (Figure 22.4.2). The bladder lies anterior to the uterus in females, posterior to the pubic bone and anterior to the rectum. During late pregnancy, its capacity is reduced due to compression by the enlarging uterus, resulting in increased frequency of urination. In males, the anatomy is similar, minus the uterus, and with the addition of the prostate inferior to the bladder. The bladder is partially retroperitoneal (outside the peritoneal cavity) with its peritoneal-covered “dome” projecting into the abdomen when the bladder is distended with urine.
Interactive Element
The bladder is a highly distensible organ because the middle layer is comprised of irregular crisscrossing bands of smooth muscle collectively called the detrusor muscle . The interior surface is made of transitional cellular epithelium that is structurally suited for the large volume fluctuations of the bladder. When empty, it resembles columnar epithelia, but when stretched, it “transitions” (hence the name) to a squamous appearance (see Figure 22.4.2). Volumes in adults can range from nearly zero to 500–600 mL. The detrusor muscle contracts with significant force in the young. The bladder’s strength diminishes with age, but voluntary contractions of abdominal skeletal muscles can increase intra-abdominal pressure to promote more forceful bladder emptying. Such voluntary contraction is also used in forceful defecation and childbirth.
Urethra
The urethra transports urine from the bladder to the outside of the body for disposal. The urethra is the only urologic organ that shows any significant anatomic difference between males and females; all other urine transport structures are identical (Figure 22.4.3) The reproductive organs will be discussed in the next chapter for both male and female. The urinary structures are discussed in this chapter with more details to in the following paragraphs.
The urethra in both males and females begins inferior and central to the two ureteral openings forming the three points of a triangular-shaped area at the base of the bladder called the trigone (Greek tri- = “triangle” and the root of the word “trigonometry”). The urethra tracks posterior and inferior to the pubic symphysis (see Figure 22.4.3). In both males and females, the proximal urethra is lined by transitional epithelium, whereas the terminal portion is a nonkeratinized, stratified squamous epithelium. In the male, pseudostratified columnar epithelium lines the urethra between these two cell types. Voiding is regulated by an involuntary autonomic nervous system-controlled internal urinary sphincter , consisting of smooth muscle and voluntary skeletal muscle that forms the external urinary sphincter below it.
Female Urethra
The external urethral orifice is embedded in the anterior vaginal wall inferior to the clitoris, superior to the vaginal opening (introitus), and medial to the labia minora. Its short length, about 4 cm, is less of a barrier to fecal bacteria than the longer male urethra and the best explanation for the greater incidence of UTI in women. Voluntary control of the external urethral sphincter is a function of the pudendal nerve. It arises in the sacral region of the spinal cord, traveling via the S2–S4 nerves of the sacral plexus.
Male Urethra
The male urethra passes through the prostate gland immediately inferior to the bladder before passing below the pubic symphysis (see Figure 22.4.3b). The length of the male urethra varies between men but averages 20 cm in length. It is divided into four regions: the preprostatic urethra , the prostatic urethra , the membranous urethra , and the spongy or penile urethra . The preprostatic urethra is very short and incorporated into the bladder wall. The prostatic urethra passes through the prostate gland. During sexual intercourse, it receives sperm via the ejaculatory ducts and secretions from the seminal vesicles. Paired Cowper’s glands (bulbourethral glands) produce and secrete mucus into the urethra to buffer urethral pH during sexual stimulation. The mucus neutralizes the usually acidic environment and lubricates the urethra, decreasing the resistance to ejaculation. The membranous urethra passes through the deep muscles of the perineum, where it is invested by the overlying urethral sphincters. The spongy urethra exits at the tip (external urethral orifice) of the penis after passing through the corpus spongiosum. Mucous glands are found along much of the length of the urethra and protect the urethra from extremes of urine pH. Innervation is the same in both males and females.
Micturition Reflex
Micturition is a less-often used, but proper term for urination or voiding. It results from an interplay of involuntary and voluntary actions by the internal and external urethral sphincters. When bladder volume reaches about 150 mL, an urge to void is sensed but is easily overridden. Voluntary control of urination relies on consciously preventing relaxation of the external urethral sphincter to maintain urinary continence. As the bladder fills, subsequent urges become harder to ignore. Ultimately, voluntary constraint fails with resulting incontinence , which will occur as bladder volume approaches 300 to 400 mL.
Nerves involved in the control of urination include the hypogastric, pelvic, and pudendal. Normal micturition is a result of stretch receptors in the bladder wall that transmit nerve impulses to the sacral region of the spinal cord to generate a spinal reflex. The resulting parasympathetic neural outflow causes contraction of the detrusor muscle and relaxation of the involuntary internal urethral sphincter. At the same time, the spinal cord inhibits somatic motor neurons, resulting in the relaxation of the skeletal muscle of the external urethral sphincter. The micturition reflex is active in infants but with maturity, children learn to override the reflex by asserting external sphincter control, thereby delaying voiding (potty training). This reflex may be preserved even in the face of spinal cord injury that results in paraplegia or quadriplegia. However, relaxation of the external sphincter may not be possible in all cases, and therefore, periodic catheterization may be necessary for bladder emptying.
Concept Review
The ureters are retroperitoneal and lead from the renal pelvis of the kidney to the trigone area at the base of the bladder. A thick muscular wall consisting of longitudinal and circular smooth muscle helps move urine toward the bladder by way of peristaltic contractions.The bladder is largely retroperitoneal and can hold up to 500–600 mL urine. The urethra is the only urinary structure that differs significantly between males and females. This is due to the dual role of the male urethra in transporting both urine and semen. The urethra arises from the trigone area at the base of the bladder. Urination is controlled by an involuntary internal sphincter of smooth muscle and a voluntary external sphincter of skeletal muscle. The shorter female urethra contributes to the higher incidence of bladder infections in females. The male urethra receives secretions from the prostate gland, Cowper’s gland, and seminal vesicles as well as sperm. Micturition is the process of voiding the urine and involves both involuntary and voluntary actions. Voluntary control of micturition requires a mature and intact sacral micturition center. It also requires an intact spinal cord. Loss of control of micturition is called incontinence and results in voiding when the bladder contains about 250 mL urine.
Review Questions
Q. Peristaltic contractions occur in the ________.
A. urethra
B. bladder
C. ureters
D. urethra, bladder, and ureters
- Answer
-
Answer: C
Q. What is the epithelium of the urinary bladder?
A. stratified squamous
B. transitional
C. stratified cuboidal
D. simple columnar
- Answer
-
Answer: B
Q. Which part of the urinary system is not completely retroperitoneal?
A. kidneys
B. ureters
C. bladder
D. nephrons
- Answer
-
Answer: C
Critical Thinking Questions
Q. Why are females more likely to contract bladder infections than males?
- Answer
-
A. The longer urethra of males means bacteria must travel farther to the bladder to cause an infection.
Q. How can the bladder fill with urine while we sleep? Gravity is not helping this process.
- Answer
-
A. Peristalsis will move the urine to the bladder, without the effect of gravity. The one-way valve allows the urine to enter the bladder but prevents reflux back into the ureter. Thus, the bladder fills up with urine over night.
Glossary
- anatomical sphincter
- smooth or skeletal muscle surrounding the lumen of a vessel or hollow organ that can restrict flow when contracted
- detrusor muscle
- smooth muscle in the bladder wall; fibers run in all directions to reduce the size of the organ when emptying it of urine
- external urinary sphincter
- skeletal muscle; must be relaxed consciously to void urine
- internal urinary sphincter
- smooth muscle at the juncture of the bladder and urethra; relaxes as the bladder fills to allow urine into the urethra
- incontinence
- loss of ability to control micturition
- micturition
- also called urination or voiding
- physiological sphincter
- sphincter consisting of circular smooth muscle indistinguishable from adjacent muscle but possessing differential innervations, permitting its function as a sphincter; structurally weak
- retroperitoneal
- outside the peritoneal cavity; in the case of the kidney and ureters, between the parietal peritoneum and the abdominal wall
- sacral micturition center
- group of neurons in the sacral region of the spinal cord that controls urination; acts reflexively unless its action is modified by higher brain centers to allow voluntary urination
- trigone
- area at the base of the bladder marked by the two ureters in the posterior–lateral aspect and the urethral orifice in the anterior aspect oriented like points on a triangle
- urethra
- transports urine from the bladder to the outside environment
Contributors and Attributions
-
OpenStax Anatomy & Physiology (CC BY 4.0). Access for free at https://openstax.org/books/anatomy-and-physiology


17.1: Sound
Learning Objectives
By the end of this section, you will be able to:
- Define sound and hearing.
- Describe sound as a longitudinal wave.
Sound can be used as a familiar illustration of waves. Because hearing is one of our most important senses, it is interesting to see how the physical properties of sound correspond to our perceptions of it. Hearing is the perception of sound, just as vision is the perception of visible light. But sound has important applications beyond hearing. Ultrasound, for example, is not heard but can be employed to form medical images and is also used in treatment.
The physical phenomenon of sound is defined to be a disturbance of matter that is transmitted from its source outward. Sound is a wave. On the atomic scale, it is a disturbance of atoms that is far more ordered than their thermal motions. In many instances, sound is a periodic wave, and the atoms undergo simple harmonic motion. In this text, we shall explore such periodic sound waves.
A vibrating string produces a sound wave as illustrated in Figures \(\PageIndex{2}\)-\(\PageIndex{4}\). As the string oscillates back and forth, it transfers energy to the air, mostly as thermal energy created by turbulence. But a small part of the string’s energy goes into compressing and expanding the surrounding air, creating slightly higher and lower local pressures. These compressions (high pressure regions) and rarefactions (low pressure regions) move out as longitudinal pressure waves having the same frequency as the string—they are the disturbance that is a sound wave. (Sound waves in air and most fluids are longitudinal, because fluids have almost no shear strength. In solids, sound waves can be both transverse and longitudinal.) Figure \(\PageIndex{4}\) shows a graph of gauge pressure versus distance from the vibrating string.
The amplitude of a sound wave decreases with distance from its source, because the energy of the wave is spread over a larger and larger area. But it is also absorbed by objects, such as the eardrum in Figure \(\PageIndex{5}\), and converted to thermal energy by the viscosity of air. In addition, during each compression a little heat transfers to the air and during each rarefaction even less heat transfers from the air, so that the heat transfer reduces the organized disturbance into random thermal motions. (These processes can be viewed as a manifestation of the second law of thermodynamics presented in Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency .) Whether the heat transfer from compression to rarefaction is significant depends on how far apart they are—that is, it depends on wavelength. Wavelength, frequency, amplitude, and speed of propagation are important for sound, as they are for all waves.
PHET EXPLORATIONS: WAVE INTERFERENCE
Make waves with a dripping faucet, audio speaker, or laser! Add a second source or a pair of slits to create an interference pattern.
Summary
- Sound is a disturbance of matter that is transmitted from its source outward.
- Sound is one type of wave.
- Hearing is the perception of sound.
Glossary
- sound
- a disturbance of matter that is transmitted from its source outward
- hearing
- the perception of sound