6.6 Surface Integrals

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

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 r. Each choice of u and v in the parameter domain gives a point on the surface, just as each choice of a parameter t gives a point on a parameterized curve. The entire surface is created by making all possible choices of u and v over the parameter domain.

Example 6.58

Parameterizing a Cylinder

Describe surface S parameterized by

$$\text{r}\left(u,v\right)=\langle \text{cos}u,\text{sin}u,v\rangle ,\text{−}\infty <u<\infty ,\text{−}\infty <v<\infty .$$

Solution

To get an idea of the shape of the surface, we first plot some points. Since the parameter domain is all of ${ℝ}^2,$ we can choose any value for u and v and plot the corresponding point. If $u=v=0,$ then $\text{r}\left(0,0\right)=\langle 1,0,0\rangle ,$ so point (1, 0, 0) is on S. Similarly, points $\text{r}\left(\pi ,2\right)=\left(\mathrm{-1},0,2\right)$ and $\text{r}\left(\frac{\pi }{2},4\right)=\left(0,1,4\right)$ 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.

First, suppose that u is a constant K. Then the curve traced out by the parameterization is $\langle \text{cos}K,\text{sin}K,v\rangle ,$ which gives a vertical line that goes through point $\left(\text{cos}K,\text{sin}K,v\right)$ in the xy-plane.

Now suppose that v is a constant K. Then the curve traced out by the parameterization is $\langle \text{cos}u,\text{sin}u,K\rangle ,$ 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 6.57).

Figure 6.57 (a) Lines $\langle \text{cos}K,\text{sin}K,v\rangle$ for $K=0,\frac{\pi }{2},\pi ,\text{and}\frac{3\pi }{2}.$ (b) Circles $\langle \text{cos}u,\text{sin}u,K\rangle$ for $K=\mathrm{-2},\mathrm{-1},1,\text{and}2.$ (c) The lines and circles together. As u and v vary, they describe a cylinder.

Notice that if $x=\text{cos}u$ and $y=\text{sin}u,$ 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 $\langle \text{cos}u,\text{sin}u,k\rangle$ for some k, and therefore each point on the cylinder is contained in the parameterized surface (Figure 6.58).

Figure 6.58 Cylinder $x^2+y^2=r^2$ has parameterization $\text{r}\left(u,v\right)=\langle r\text{cos}u,r\text{sin}u,v\rangle ,$ $0\le u\le 2\pi ,\text{−}\infty <v<\infty .$

Analysis

Notice that if we change the parameter domain, we could get a different surface. For example, if we restricted the domain to $0\le u\le \pi ,0<v<6,$ then the surface would be a half-cylinder of height 6.

It follows from Example 6.58 that we can parameterize all cylinders of the form $x^2+y^2=R^2.$ If S is a cylinder given by equation $x^2+y^2=R^2,$ then a parameterization of S is

We can also find different types of surfaces given their parameterization, or we can find a parameterization when we are given a surface.

Example 6.59

Describing a Surface

Describe surface S parameterized by

$$\text{r}(u,v)=\langle u\text{cos}v,u\text{sin}v,u^2\rangle ,0\le u<\infty ,0\le v<2\pi .$$

Solution

Notice that if u is held constant, then the resulting curve is a circle of radius u in plane $z=u^2.$ 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

$$\begin{array}{cc}\hfill x^2+y^2& ={\left(u\text{cos}v\right)}^2+{\left(u\text{sin}v\right)}^2\hfill \\ & =u^2{\text{cos}}^{2}v+u^2{\text{sin}}^{2}v\hfill \\ & =u^2\hfill \\ & =z.\hfill \end{array}$$

Therefore, the surface is elliptic paraboloid $x^2+y^2=z$ (Figure 6.59).

Figure 6.59 (a) Circles arise from holding u constant; the vertical parabolas arise from holding v constant. (b) An elliptic paraboloid results from all choices of u and v in the parameter domain.

Example 6.60

Finding a Parameterization

Give a parameterization of the cone $x^2+y^2=z^2$ lying on or above the plane $z=\mathrm{-2}.$

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 $\left(u\text{cos}v,u\text{sin}v,u\right)$ for angle v. Hence, a parameterization of the cone is $\text{r}(u,v)=\langle u\text{cos}v,u\text{sin}v,u\rangle .$ Since we are not interested in the entire cone, only the portion on or above plane $z=\mathrm{-2},$ the parameter domain is given by $\mathrm{-2}\le u<\infty ,0\le v<2\pi$ (Figure 6.60).

Figure 6.60 Cone $x^2+y^2=z^2$ has parameterization $r(u,v)=\langle u\text{cos}v,u\text{sin}v,u\rangle ,\text{−}\infty <u<\infty ,0\le v\le 2\pi .$

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

The idea of this parameterization is that as $\varphi$ sweeps downward from the positive z-axis, a circle of radius $\rho \text{sin}\varphi$ is traced out by letting $\theta$ run from 0 to $2\pi .$ To see this, let $\varphi$ be fixed. Then

This results in the desired circle (Figure 6.61).

Figure 6.61 The sphere of radius $\rho$ has parameterization $\text{r}\left(\varphi ,\theta \right)=\langle \rho \text{cos}\theta \text{sin}\varphi ,\rho \text{sin}\theta \text{sin}\varphi ,\rho \text{cos}\varphi \rangle ,$ $0\le \theta \le 2\pi ,0\le \varphi \le \pi .$

Finally, to parameterize the graph of a two-variable function, we first let $z=f\left(x,y\right)$ be a function of two variables. The simplest parameterization of the graph of $f$ is $\text{r}(x,y)=\langle x,y,f\left(x,y\right)\rangle ,$ where x and y vary over the domain of $f$ (Figure 6.62). For example, the graph of $f(x,y)=x^{2}y$ can be parameterized by $\text{r}(x,y)=\langle x,y,x^{2}y\rangle ,$ 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 $\left[1,3\right]\times \left[2,5\right]$—then we can restrict the parameter domain to give this piece of the surface:

Similarly, if S is a surface given by equation $x=g\left(y,z\right)$ or equation $y=h(x,z),$ then a parameterization of S is

$\text{r}(y,z)=\langle g\left(y,z\right),y,z\rangle$ or $\text{r}(x,z)=\langle x,h\left(x,z\right),z\rangle ,$ respectively. For example, the graph of paraboloid $2y=x^2+z^2$ can be parameterized by $\text{r}(x,z)=\langle x,\frac{x^2+z^2}{2},z\rangle ,0\le x<\infty ,0\le z<\infty .$ Notice that we do not need to vary over the entire domain of y because x and z are squared.

Figure 6.62 The simplest parameterization of the graph of a function is $\text{r}(x,y)=\langle x,y,f\left(x,y\right)\rangle .$

Let’s now generalize the notions of smoothness and regularity to a parametric surface. Recall that curve parameterization $\text{r}(t),a\le t\le b$ is regular if $\text{r}\prime (t)\ne 0$ for all t in $[a,b].$ For a curve, this condition ensures that the image of r really is a curve, and not just a point. For example, consider curve parameterization $\text{r}(t)=\langle 1,2\rangle ,0\le t\le 5.$ The image of this parameterization is simply point $(1,2),$ which is not a curve. Notice also that $\text{r}\prime (t)=0.$ 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 for surfaces so that a surface parameterization really does trace out a surface. To motivate the definition of regularity of a surface parameterization, consider parameterization

Although this parameterization appears to be the parameterization of a surface, notice that the image is actually a line (Figure 6.63). How could we avoid parameterizations such as this? Parameterizations that do not give an actual surface? Notice that ${\text{r}}_u=\langle 0,0,0\rangle$ and ${\text{r}}_v=\langle 0,\text{−}\text{sin}v,0\rangle ,$ and the corresponding cross product is zero. The analog of the condition $\text{r}\prime (t)=\mathbf{0}$ is that ${\text{r}}_u\times {\text{r}}_v$ is not zero for any point $(u,v)$ in the parameter domain, which is a regular parameterization.

Figure 6.63 The image of parameterization $\text{r}(u,v)=\langle 0,\text{cos}v,1\rangle ,0\le u\le 1,0\le v\le \pi$ is a line.

If parameterization r is regular, then the image of r is a two-dimensional object, as a surface should be. Throughout this chapter, parameterizations $\text{r}(u,v)=\langle x(u,v),y(u,v),z(u,v)\rangle$ are assumed to be regular.

Recall that curve parameterization $\text{r}(t),a\le t\le b$ is smooth if $\text{r}\prime (t)$ is continuous and $\text{r}\prime (t)\ne \mathbf{0}$ 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.

A surface may also be piecewise smooth if it has smooth faces but also has locations where the directional derivatives do not exist.

Example 6.61

Identifying Smooth and Nonsmooth Surfaces

Which of the figures in Figure 6.64 is smooth?

Figure 6.64 (a) This surface is smooth. (b) This surface is piecewise smooth.

Solution

The surface in Figure 6.64(a) can be parameterized by

$$\text{r}(u,v)=\langle (2+\text{cos}v)\text{cos}u,(2+\text{cos}v)\text{sin}u,\text{sin}v\rangle ,0\le u<2\pi ,0\le v<2\pi$$

(we can use technology to verify). Notice that vectors

$${\text{r}}_u=\langle \text{−}(2+\text{cos}v)\text{sin}u,(2+\text{cos}v)\text{cos}u,0\rangle \text{and}{\text{r}}_v=\langle \text{−}\text{sin}v\text{cos}u,\text{−}\text{sin}v\text{sin}u,\text{cos}v\rangle$$

exist for any choice of u and v in the parameter domain, and

$$\begin{array}{cc}\hfill {\text{r}}_u\times {\text{r}}_v& =\left|\begin{array}{ccc}\hfill \text{i}\hfill & \hfill \text{j}\hfill & \hfill \text{k}\hfill \\ \hfill \text{−}(2+\text{cos}v)\text{sin}u\hfill & \hfill (2+\text{cos}v)\text{cos}u\hfill & \hfill 0\hfill \\ \hfill \text{−}\text{sin}v\text{cos}u\hfill & \hfill \text{−}\text{sin}v\text{sin}u\hfill & \hfill \text{cos}v\hfill \end{array}\right|\hfill \\ & =\left[(2+\text{cos}v)\text{cos}u\text{cos}v\right]\text{i}+\left[(2+\text{cos}v)\text{sin}u\text{cos}v\right]\text{j}\hfill \\ & +\left[(2+\text{cos}v)\text{sin}v{\text{sin}}^{2}u+(2+\text{cos}v)\text{sin}v{\text{cos}}^{2}u\right]\text{k}\hfill \\ & =\left[(2+\text{cos}v)\text{cos}u\text{cos}v\right]\text{i}+\left[(2+\text{cos}v)\text{sin}u\text{cos}v\right]\text{j}+\left[(2+\text{cos}v)\text{sin}v\right]\text{k}.\hfill \end{array}$$

The k component of this vector is zero only if $v=0$ or $v=\pi .$ If $v=0$ or $v=\pi ,$ then the only choices for u that make the j component zero are $u=0$ or $u=\pi .$ But, these choices of u do not make the i component zero. Therefore, ${\text{r}}_u\times {\text{r}}_v$ 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 6.64(b), 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 five smooth faces, and thus this surface is piecewise smooth.

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 $\text{r}\left(u,v\right)=\langle x\left(u,v\right),y\left(u,v\right),z\left(u,v\right)\rangle$ over some parameter domain D. We assume here and throughout that the surface parameterization $\text{r}\left(u,v\right)=\langle x\left(u,v\right),y\left(u,v\right),z\left(u,v\right)\rangle$ 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 $\text{Δ}u$ and vertical length $\text{Δ}v.$ 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 surface S into pieces $S_{ij}.$ 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 $\text{r}\left(u,v\right).$ Without loss of generality, we assume that $P_{ij}$ is located at the corner of two grid curves, as in Figure 6.65. If we think of r as a mapping from the uv-plane to ${ℝ}^3,$ the grid curves are the image of the grid lines under 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 $\text{r}(u_i,v)$ and $\text{r}(u,v_j),$ and these curves intersect at point $P_{ij}.$

Figure 6.65 Grid lines on a parameter domain correspond to grid curves on a surface.

Now consider the vectors that are tangent to these grid curves. For grid curve $\text{r}(u_i,v),$ the tangent vector at $P_{ij}$ is

For grid curve $\text{r}(u,v_j),$ the tangent vector at $P_{ij}$ is

If vector $\text{N}={\text{t}}_u\left(P_{ij}\right)\times {\text{t}}_v\left(P_{ij}\right)$ exists and is not zero, then the tangent plane at $P_{ij}$ exists (Figure 6.66). If piece $S_{ij}$ is small enough, then the tangent plane at point $P_{ij}$ is a good approximation of piece $S_{ij}.$

Figure 6.66 If the cross product of vectors ${\text{t}}_u$ and ${\text{t}}_v$ exists, then there is a tangent plane.

The tangent plane at $P_{ij}$ contains vectors ${\text{t}}_u\left(P_{ij}\right)$ and ${\text{t}}_v\left(P_{ij}\right),$ and therefore the parallelogram spanned by ${\text{t}}_u\left(P_{ij}\right)$ and ${\text{t}}_v\left(P_{ij}\right)$ is in the tangent plane. Since the original rectangle in the uv-plane corresponding to $S_{ij}$ has width $\text{Δ}u$ and length $\text{Δ}v,$ the parallelogram that we use to approximate $S_{ij}$ is the parallelogram spanned by $\text{Δ}u{\text{t}}_u\left(P_{ij}\right)$ and $\text{Δ}v{\text{t}}_v\left(P_{ij}\right).$ In other words, we scale the tangent vectors by the constants $\text{Δ}u$ and $\text{Δ}v$ 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

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 6.67).

Figure 6.67 The parallelogram spanned by ${\text{t}}_u$ and ${\text{t}}_v$ approximates the piece of surface $S_{ij}.$

Example 6.62

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 6.18, we need a parameterization. We assume this cone is in ${ℝ}^3$ with its vertex at the origin (Figure 6.68). 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 $k=\text{tan}\alpha .$ For a height value v with $0\le v\le h,$ the radius of the circle formed by intersecting the cone with plane $z=v$ is $kv.$ Therefore, a parameterization of this cone is

$$\text{s}(u,v)=\langle kv\text{cos}u,kv\text{sin}u,v\rangle ,0\le u<2\pi ,0\le v\le h.$$

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.

Figure 6.68 The right circular cone with radius r = kh and height h has parameterization $\text{s}(u,v)=\langle kv\text{cos}u,kv\text{sin}u,v\rangle ,0\le u<2\pi ,0\le v\le h.$

With a parameterization in hand, we can calculate the surface area of the cone using Equation 6.18. The tangent vectors are ${\text{t}}_u=\langle \text{−}kv\text{sin}u,kv\text{cos}u,0\rangle$ and ${\text{t}}_v=\langle k\text{cos}u,k\text{sin}u,1\rangle .$ Therefore,

$$\begin{array}{cc}\hfill {\text{t}}_u\times {\text{t}}_v& =\left|\begin{array}{ccc}\hfill \text{i}\hfill & \hfill \text{j}\hfill & \hfill \text{k}\hfill \\ \hfill \text{−}kv\text{sin}u\hfill & \hfill kv\text{cos}u\hfill & \hfill 0\hfill \\ \hfill k\text{cos}u\hfill & \hfill k\text{sin}u\hfill & \hfill 1\hfill \end{array}\right|\hfill \\ & =\langle kv\text{cos}u,kv\text{sin}u,\text{−}{k}^{2}v{\text{sin}}^{2}u-k^{2}v{\text{cos}}^{2}u\rangle \hfill \\ & =\langle kv\text{cos}u,kv\text{sin}u,\text{−}{k}^{2}v\rangle .\hfill \end{array}$$

The magnitude of this vector is

$$\begin{array}{cc}\hfill \Vert \langle kv\text{cos}u,kv\text{sin}u,\text{−}{k}^{2}v\rangle \Vert & =\sqrt{k^2{v}^2{\text{cos}}^{2}u+k^2{v}^2{\text{sin}}^{2}u+k^4{v}^2}\hfill \\ & =\sqrt{k^2{v}^2+k^4{v}^2}\hfill \\ & =kv\sqrt{1+k^2}.\hfill \end{array}$$

By Equation 6.18, the surface area of the cone is

$$\begin{array}{cc}\hfill {\displaystyle {\iint }_D\Vert {\text{t}}_u\times {\text{t}}_v\Vert dA}& ={\displaystyle {\int }_0^h{\displaystyle {\int }_0^{2\pi }kv\sqrt{1+k^2}}dudv}\hfill \\ & =2\pi k\sqrt{1+k^2}{\displaystyle {\int }_0^{h}vdv}\hfill \\ & =2\pi k\sqrt{1+k^2}{\left[\frac{v^2}{2}\right]}_0^h\hfill \\ & =\pi k{h}^2\sqrt{1+k^2}.\hfill \end{array}$$

Since $k=\text{tan}\alpha =r\text{/}h,$

$$\begin{array}{cc}\hfill \pi k{h}^2\sqrt{1+k^2}& =\pi \frac{r}{h}{h}^2\sqrt{1+\frac{r^2}{h^2}}\hfill \\ & =\pi rh\sqrt{1+\frac{r^2}{h^2}}\hfill \\ & =\pi r\sqrt{h^2+h^2\left(\frac{r^2}{h^2}\right)}\hfill \\ & =\pi r\sqrt{h^2+r^2}.\hfill \end{array}$$

Therefore, the lateral surface area of the cone is $\pi r\sqrt{h^2+r^2}.$

Analysis

The 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.