Surface Integrals: Learn It 1

Giselle Miller

Surface Integrals: Learn It 1

Understand and find parametric representations of surfaces
Calculate surface integrals of scalar functions
Calculate surface integrals of vector fields
Use surface integrals to solve real-world problems

Parametric Surfaces

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.

Just as with line integrals, there are two kinds of surface integrals: surface integrals of scalar-valued functions and surface integrals of vector fields. Before we can integrate over a surface, we need to understand how to describe surfaces mathematically using parameterizations.

Parameterizing Curves

To calculate a line integral over curve [latex]C[/latex], we first parameterize [latex]C[/latex]. A parametric curve uses one parameter [latex]t[/latex] to trace out a one-dimensional path.

To calculate a surface integral over surface [latex]S[/latex], we need to parameterize [latex]S[/latex]. A parametric surface uses two parameters because a surface is two-dimensional.

A parametric surface is given by:

[latex]\mathbf{r}(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle[/latex]

Notice that this parameterization involves two parameters, [latex]u[/latex] and [latex]v[/latex], because a surface is two-dimensional, and therefore two variables are needed to trace out the surface.

The parameters [latex]u[/latex] and [latex]v[/latex] vary over a region called the parameter domain (or parameter space)—the set of points in the [latex]uv[/latex]-plane that can be substituted into [latex]\mathbf{r}[/latex]. Each choice of [latex]u[/latex] and [latex]v[/latex] in the parameter domain gives a point on the surface, just as each choice of parameter [latex]t[/latex] gives a point on a parameterized curve. The entire surface is created by making all possible choices of [latex]u[/latex] and [latex]v[/latex] over the parameter domain.

parameter domain

Given a parameterization of surface [latex]{\bf{r}}(u,v)=\langle{x}(u,v),y(u,v),z(u,v)\rangle[/latex], the parameter domain of the parameterization is the set of points in the [latex]uv[/latex]-plane that can be substituted into [latex]{\bf{r}}[/latex].

Describe the surface [latex]S[/latex] parameterized by

[latex]\mathbf{r}(u,v) = \langle \cos u, \sin u, v \rangle, -\infty < u < \infty, -\infty < v < \infty[/latex]

To get an idea of the shape of the surface, we first plot some points. Since the parameter domain is all of [latex]\mathbb{R}^2[/latex], we can choose any value for [latex]u[/latex] and [latex]v[/latex] and plot the corresponding point. If [latex]u=v=0[/latex], then [latex]{\bf{r}}(0,0)\langle1,0,0\rangle[/latex], so point [latex](1, 0, 0)[/latex] is on [latex]S[/latex]. Similarly, points [latex]{\bf{r}}(\pi,2)=\langle-1,0,2\rangle[/latex] and [latex]{\bf{r}}\left(\frac{\pi}2,4\right)=(0,1,4)[/latex] are on [latex]S[/latex].

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 [latex]S[/latex], we visualize two families of curves that lie on [latex]S[/latex]. In the first family of curves we hold [latex]u[/latex] constant; in the second family of curves we hold [latex]v[/latex] constant. This allows us to build a “skeleton” of the surface, thereby getting an idea of its shape.

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

Now suppose that [latex]v[/latex] is a constant [latex]K[/latex]. Then the curve traced out by the parameterization is [latex]\langle\cos{K},\sin{K},v\rangle[/latex], which gives a circle in plane [latex]z=K[/latex] with radius 1 and center [latex](0, 0, K)[/latex].

If [latex]u[/latex] is held constant, then we get vertical lines; if [latex]v[/latex] 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 [latex]x^{2}+y^{2}=1[/latex] (Figure 1).

Three diagrams in three dimensions. The first shows vertical lines around the origin. The second shows parallel circles all with center at the origin and radius of 1. The third shows the lines and circle. Together, they form the skeleton of a cylinder. — Figure 1. (a) Lines [latex]\langle\cos{K},\sin{K},v\rangle[/latex] for [latex]K = 0, \frac{\pi}2, π,[/latex] and [latex]\frac{3\pi}2[/latex]. (b) Circles [latex]\langle\cos{u},\sin{u},K\rangle[/latex] for [latex]K = −2, −1, 1,[/latex] and [latex]2[/latex]. (c) The lines and circles together. As [latex]u[/latex] and [latex]v[/latex] vary, they describe a cylinder.

Notice that if [latex]x=\cos{u}[/latex] and [latex]y=\sin{u}[/latex], then [latex]x^{2}+y^{2}=1[/latex], so points from [latex]S[/latex] do indeed lie on the cylinder. Conversely, each point on the cylinder is contained in some circle [latex]\langle\cos{K},\sin{K},v\rangle[/latex] for some [latex]k[/latex], and therefore each point on the cylinder is contained in the parameterized surface (Figure 2).

An image of a vertical cylinder in three dimensions with the center of its circular base located on the z axis. — Figure 2. Cylinder [latex]x^{2}+y^{2}=r^{2}[/latex] has parameterization [latex]{\bf{r}}(u,v)=\langle{r}\cos{u},r\sin{u},v\rangle[/latex], [latex]0\leq{u}\leq2\pi,-\infty<v<\infty[/latex].

[latex]\\[/latex]
Analysis
[latex]\\[/latex]
Notice that if we change the parameter domain, we could get a different surface. For example, if we restricted the domain to [latex]0\leq{u}\leq\pi, 0 < v <6[/latex], then the surface would be a half-cylinder of height 6.

Parameterizing Cylinders

It follows from the example that we can parameterize all cylinders of the form [latex]x^2 + y^2 = R^2[/latex]. If [latex]S[/latex] is a cylinder given by equation [latex]x^2 + y^2 = R^2[/latex], then a parameterization of [latex]S[/latex] is:

[latex]\mathbf{r}(u,v) = \langle R\cos u, R\sin u, v \rangle, \quad 0 \leq u \leq 2\pi, \quad -\infty < v < \infty[/latex]

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

Describe the surface [latex]S[/latex] parameterized by [latex]{\bf{r}}(u,v)=\langle{u}\cos{v},u\sin{v},u^2\rangle,0\le{u}\le\infty,0\le{v}<2\pi[/latex].

Give a parameterization of the cone [latex]x^{2}+y^{2}=z^{2}[/latex] lying on or above the plane [latex]z=-2[/latex].

We have discussed parameterizations of various surfaces, but two important types of surfaces need a separate discussion: spheres and graphs of two-variable functions.

Parameterizing Spheres

To parameterize a sphere, it’s easiest to use spherical coordinates. The sphere of radius [latex]\rho[/latex] centered at the origin is given by:

[latex]\mathbf{r}(\phi,\theta) = \langle \rho\cos\theta\sin\phi, \rho\sin\theta\sin\phi, \rho\cos\phi \rangle, \quad 0 \leq \theta \leq 2\pi, \quad 0 \leq \phi \leq \pi[/latex]

The idea is that as [latex]\phi[/latex] sweeps downward from the positive [latex]z[/latex]-axis, a circle of radius [latex]\rho\sin\phi[/latex] is traced out by letting [latex]\theta[/latex] run from [latex]0[/latex] to [latex]2\pi[/latex]. To see this, let [latex]\phi[/latex] be fixed. Then:

[latex]\begin{aligned} x^2 + y^2 &= (\rho\cos\theta\sin\phi)^2 + (\rho\sin\theta\sin\phi)^2 \\ &= \rho^2\sin^2\phi(\cos^2\theta + \sin^2\theta) \\ &= \rho^2\sin^2\phi \ &= (\rho\sin\phi)^2 \end{aligned}[/latex]

This gives the desired circle (Figure 5).

A three-dimensional diagram of the sphere of radius rho. — Figure 5. The sphere of radius [latex]\rho[/latex] has parameterization [latex]{\bf{r}}(\phi,\theta)=\langle\rho\cos\theta\sin\phi,\rho\sin\theta\sin\phi,\rho\cos\phi\rangle[/latex],[latex] 0\leq\theta\leq2\pi, 0\leq\phi\leq\pi[/latex].

Parameterizing Graphs of Functions

To parameterize the graph of a two-variable function, let [latex]z = f(x,y)[/latex] be a function of two variables. The simplest parameterization of the graph of [latex]f[/latex] is:

[latex]\mathbf{r}(x,y) = \langle x, y, f(x,y) \rangle[/latex]

where [latex]x[/latex] and [latex]y[/latex] vary over the domain of [latex]f[/latex] (Figure 6).

For example, the graph of [latex]f(x,y) = x^2y[/latex] can be parameterized by [latex]\mathbf{r}(x,y) = \langle x, y, x^2y \rangle[/latex], where the parameters [latex]x[/latex] and [latex]y[/latex] vary over the domain of [latex]f[/latex].

If you only care about a piece of the graph of [latex]f[/latex]—say, the piece over rectangle [latex][1,3] \times [2,5][/latex]—then you can restrict the parameter domain:

[latex]\mathbf{r}(x,y) = \langle x, y, x^2y \rangle, \quad 1 \leq x \leq 3, \quad 2 \leq y \leq 5[/latex]

Similarly, if [latex]S[/latex] is a surface given by [latex]x = g(y,z)[/latex] or [latex]y = h(x,z)[/latex], then a parameterization of [latex]S[/latex] is:

[latex]\mathbf{r}(y,z) = \langle g(y,z), y, z \rangle[/latex] or [latex]\mathbf{r}(x,z) = \langle x, h(x,z), z \rangle[/latex]

For example, the graph of paraboloid [latex]2y = x^2 + z^2[/latex] can be parameterized by:

[latex]\mathbf{r}(x,z) = \left\langle x, \frac{x^2 + z^2}{2}, z \right\rangle, \quad 0 \leq x \leq \infty, \quad 0 \leq z \leq \infty[/latex]

Notice that we don’t need to vary over the entire domain of [latex]y[/latex] because [latex]x[/latex] and [latex]z[/latex] are squared.

A three-dimensional diagram of a surface z = f(x,y) above its mapping in the two-dimensional x,y plane. The point (x,y) in the plane corresponds to the point z = f(x,y) on the surface. — Figure 6. The simplest parameterization of the graph of a function is [latex]{\bf{r}}(x,y)=\langle{x},y,f(x,y)\rangle[/latex].

Smooth and Regular Parameterizations

Let’s generalize the notions of smoothness and regularity from curves to parametric surfaces.

Regular Parameterizations of Curves

Curve parameterization [latex]\mathbf{r}(t)[/latex], [latex]a \leq t \leq b[/latex], is regular if [latex]\mathbf{r}'(t) \neq \mathbf{0}[/latex] for all [latex]t[/latex] in [latex][a,b][/latex]. This condition ensures the image of [latex]\mathbf{r}[/latex] is actually a curve, not just a point.

For example, [latex]\mathbf{r}(t) = \langle 1, 2 \rangle[/latex], [latex]0 \leq t \leq 5[/latex], is not regular because its image is just the point [latex](1, 2)[/latex] and [latex]\mathbf{r}'(t) = \mathbf{0}[/latex].

We need a similar notion of regularity for surfaces to ensure a surface parameterization traces out an actual surface. Consider the parameterization:

[latex]\mathbf{r}(u,v) = \langle 0, \cos v, 1 \rangle, \quad 0 \leq u \leq 1, \quad 0 \leq v \leq \pi[/latex]

Although this appears to parameterize a surface, its image is actually just a line (Figure 7). Notice that [latex]\mathbf{r}_u = \langle 0, 0, 0 \rangle[/latex] and [latex]\mathbf{r}_v = \langle 0, -\sin v, 0 \rangle[/latex], so the cross product [latex]\mathbf{r}_u \times \mathbf{r}_v = \mathbf{0}[/latex].

The analog of the condition [latex]\mathbf{r}'(t) \neq \mathbf{0}[/latex] for curves is that [latex]\mathbf{r}_u \times \mathbf{r}_v \neq \mathbf{0}[/latex] for surfaces.

A three-dimensional diagram of a line on the x,z plane where the z component is 1, the x component is 1, and the y component exists between -1 and 1. — Figure 7. The image of parameterization [latex]{{\bf{r}}(u,v)=\langle0,\cos{v},1\rangle, \ 0\leq{u}\leq1, \ 0\leq{v}\leq\pi}[/latex] is a line.

regular parameterization

Parameterization [latex]{\bf{r}}(u,v)=\langle(x(u,v),y(u,v),z(u,v)\rangle[/latex] is a regular parameterization if [latex]{\bf{r}}_u\times{\bf{r}}_v[/latex] is not zero for point [latex](u, v)[/latex] in the parameter domain.

If parameterization [latex]\mathbf{r}[/latex] is regular, then the image of [latex]\mathbf{r}[/latex] is a two-dimensional object, as a surface should be. Throughout this module, we assume all parameterizations [latex]\mathbf{r}(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle[/latex] are regular.

Smooth Parameterizations of Curves

Curve parameterization [latex]\mathbf{r}(t)[/latex], [latex]a \leq t \leq b[/latex], is smooth if [latex]\mathbf{r}'(t)[/latex] is continuous and [latex]\mathbf{r}'(t) \neq \mathbf{0}[/latex] for all [latex]t[/latex] in [latex][a,b][/latex]. 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.

smooth parameterization

A surface parameterization [latex]{\bf{r}}(u,v)=\langle(x(u,v),y(u,v),z(u,v)\rangle[/latex] is smooth if vector [latex]{\bf{r}}_u\times{\bf{r}}_v[/latex] is not zero for any choice of [latex]u[/latex] and [latex]v[/latex] 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.

Which of the figures in Figure 8 is smooth?

Two three-dimensional figures. The first surface is smooth. It looks like a tire with a large hole in the middle. The second is piecewise smooth. It is a pyramid with a rectangular base and four sides. — Figure 8. (a) This surface is smooth. (b) This surface is piecewise smooth.