14 Multiple Integration

14.5 Surface Area

In Section 10.1 we used definite integrals to compute the arc length of plane curves of the form y=f(x). We later extended these ideas to compute the arc length of plane curves defined by parametric or polar equations.

The natural extension of the concept of “arc length over an interval” to surfaces is “surface area over a region.”

Consider the surface z=f(x,y) over a region R in the x-y plane, shown in Figure 14.5.1(a). Because of the domed shape of the surface, the surface area will be greater than that of the area of the region R. We can find this area using the same basic technique we have used over and over: we’ll make an approximation, then using limits, we’ll refine the approximation to the exact value. margin:
Figure 14.5.1: Developing a method of computing surface area. Λ

As done to find the volume under a surface or the mass of a lamina, we subdivide R into n subregions. Here we subdivide R into rectangles, as shown in the figure. One such subregion is outlined in the figure, where the rectangle has dimensions Δxi and Δyi, along with its corresponding region on the surface.

In part (b) of the figure, we zoom in on this portion of the surface. When Δxi and Δyi are small, the function is approximated well by the tangent plane at any point (xi,yi) in this subregion, which is graphed in part (b). In fact, the tangent plane approximates the function so well that in this figure, it is virtually indistinguishable from the surface itself! Therefore we can approximate the surface area Si of this region of the surface with the area Ti of the corresponding portion of the tangent plane.

This portion of the tangent plane is a parallelogram, defined by sides u and v, as shown. One of the applications of the cross product from Section 11.4 is that the area of this parallelogram is u×v. Once we can determine u and v, we can determine the area.

u is tangent to the surface in the direction of x, therefore, from Section 13.7, u is parallel to 1,0,fx(xi,yi). The x-displacement of u is Δxi, so we know that u=Δxi1,0,fx(xi,yi). Similar logic shows that v=Δyi0,1,fy(xi,yi). Thus:

surface area Si area of Ti

Note that ΔxiΔyi=ΔAi, the area of the ith subregion.

Summing up all n of the approximations to the surface area gives

surface area over Ri=1n1+[fx(xi,yi)]2+[fy(xi,yi)]2ΔAi.

Once again take a limit as all of the Δxi and Δyi shrink to 0; this leads to a double integral.

margin: Note: As before, we think of “RdS” as meaning “sum up lots of little surface areas over R.” The concept of surface area is defined here, for while we already have a notion of the area of a region in the plane, we did not yet have a solid grasp of what “the area of a surface in space” means. Λ
Definition 14.5.1 Surface Area

Let z=f(x,y) where fx and fy are continuous over a closed, bounded region R. The surface area S over R is

S =RdS

We test this definition by using it to compute surface areas of known surfaces. We start with a triangle.

Example 14.5.1 Finding the surface area of a plane over a triangle

Let f(x,y)=4-x-2y, and let R be the region in the plane bounded by x=0, y=0 and y=2-x/2, as shown in Figure 14.5.2. Find the surface area of z=f(x,y) over R. margin:
Figure 14.5.2: Finding the area of a triangle in space in Example 14.5.1. Λ

SolutionWe follow Definition 14.5.1 and start by noting fx(x,y)=-1 and fy(x,y)=-2. To define R, we use bounds 0y2-x/2 and 0x4. Therefore

S =RdS

Because the surface is a triangle, we can figure out the area using geometry. Considering the base of the triangle to be the side in the x-y plane, we find the length of the base to be 20. We can find the height using our knowledge of vectors: let u be the side in the x-z plane and let v be the side in the x-y plane. The height is then u-projvu=46/5. Geometry states that the area is thus


We affirm the validity of our formula.

It is “common knowledge” that the surface area of a sphere of radius r is 4πr2. We confirm this in the following example, which involves using our formula with polar coordinates.

Example 14.5.2 The surface area of a sphere.

Find the surface area of the sphere with radius a centered at the origin, whose top hemisphere has equation z=f(x,y)=a2-x2-y2.

SolutionWe start by computing partial derivatives and find


As our function f only defines the top upper hemisphere of the sphere, we double our surface area result to get the total area:

S =2R1+[fx(x,y)]2+[fy(x,y)]2dA

The region R that we are integrating over is the disk, centered at the origin, with radius a: x2+y2a2. Because of this region, we are likely to have greater success with our integration by converting to polar coordinates. Using the substitutions x=rcosθ, y=rsinθ, dA=rdrdθ and bounds 0θ2π and 0ra, we have: margin: Note: The inner integral in Equation (14.5.1) is an improper integral, as the integrand of 0ara2a2-r2dr is not defined at r=a. To properly evaluate this integral, one must use the techniques of Section 8.6. The reason this need arises is that the function f(x,y)=a2-x2-y2 fails the requirements of Definition 14.5.1, as fx and fy are not continuous on the boundary of the circle x2+y2=a2. The computation of the surface area is still valid. The definition makes stronger requirements than necessary in part to avoid the use of improper integration, as when fx and/or fy are not continuous, the resulting improper integral may not converge. Since the improper integral does converge in this example, the surface area is accurately computed. Λ

S =202π0a1+r2cos2θ+r2sin2θa2-r2cos2θ-r2sin2θrdrdθ
=202π0ara2a2-r2drdθ. (14.5.1)
Apply substitution u=a2-r2 and integrate the inner integral, giving

Our work confirms our previous formula.

Example 14.5.3 Finding the surface area of a cone

The general formula for a right cone with height h and base radius a, as shown in Figure 14.5.3, is margin:
Figure 14.5.3: Finding the surface area of a cone in Example 14.5.3. Λ


Find the surface area of this cone.

SolutionWe begin by computing partial derivatives.

fx(x,y)=-xhax2+y2  and  fy(x,y)=-yhax2+y2.

Since we are integrating over the disk x2+y2a2, we again use polar coordinates. Using the standard substitutions, our integrand becomes


This may look intimidating at first, but there are lots of simple simplifications to be done. It amazingly reduces to just


Our polar bounds are 0θ2π and 0ra. Thus margin: Note: Note that once again fx and fy are not continuous on the domain of f, as both are undefined at (0,0). (A similar problem occurred in the previous example.) Once again the resulting improper integral converges and the computation of the surface area is valid. Λ

S =02π0ar1aa2+h2drdθ

This matches the formula found in the back of this text.

Example 14.5.4 Finding surface area over a region

Find the area of the surface z=f(x,y)=x2-3y+3 over the region R bounded by -xyx, 0x4, as pictured in Figure 14.5.4.

SolutionIt is straightforward to compute fx(x,y)=2x and fy(x,y)=-3. Thus the surface area is described by the double integral margin:
Figure 14.5.4: Graphing the surface in Example 14.5.4. Λ


As with integrals describing arc length, double integrals describing surface area are in general hard to evaluate directly because of the square-root. This particular integral can be easily evaluated, though, with judicious choice of our order of integration.

Integrating with order dxdy requires us to evaluate 10+4x2dx. This can be done, though it involves Integration By Parts and sinh-1x. Integrating with order dydx has as its first integral 10+4x2dy, which is easy to evaluate: it is simply y10+4x2+C. So we proceed with the order dydx; the bounds are already given in the statement of the problem.

R10+4x2dA =04-xx10+4x2dydx
Apply substitution with u=10+4x2:
=13(3774-510) units2.

So while the region R over which we integrate has an area of 16 units2, the surface has a much greater area as its z-values change dramatically over R.

In practice, technology helps greatly in the evaluation of such integrals. High powered computer algebra systems can compute integrals that are difficult, or at least time consuming, by hand, and can at the least produce very accurate approximations with numerical methods. In general, just knowing how to set up the proper integrals brings one very close to being able to compute the needed value. Most of the work is actually done in just describing the region R in terms of polar or rectangular coordinates. Once this is done, technology can usually provide a good answer.

We have learned how to integrate integrals; that is, we have learned to evaluate double integrals. In the next section, we learn how to integrate double integrals — that is, we learn to evaluate triple integrals, along with learning some uses for this operation.

Exercises 14.5


Terms and Concepts

  1. 1.

    “Surface area” is analogous to what previously studied concept?

  2. 2.

    To approximate the area of a small portion of a surface, we computed the area of its              plane.

  3. 3.

    We interpret RdS as “sum up lots of little                          .”

  4. 4.

    Why is it important to know how to set up a double integral to compute surface area, even if the resulting integral is hard to evaluate?

  5. 5.

    Why do the graphs of z=f(x,y) and z=g(x,y)=f(x,y)+h, for some real number h, have the same surface area over a region R?

  6. 6.

    Let z=f(x,y) and z=g(x,y)=2f(x,y). Why is the surface area of the graph of g over a region R not twice the surface area of the graph of f over R?


In Exercises 7–10., set up the iterated integral that computes the surface area of the surface z=f(x,y) over the region R.

  1. 7.
    f(x,y)=sinxcosy; R is the rectangle with bounds 0x2π,  0y2π.
  2. 8.
    f(x,y)=1x2+y2+1; R is the disk x2+y29.
  3. 9.
    f(x,y)=x2-y2; R is the rectangle with opposite corners (-1,-1) and (1,1).
  4. 10.
    f(x,y)=1ex2+1; R is the rectangle bounded by -5x5 and 0y1.

In Exercises 11–18., find the area of the surface of z=f(x,y) over the region R.

  1. 11.

    f(x,y)=3x-7y+2; R is the rectangle with opposite corners (-1,0) and (1,3).

  2. 12.

    f(x,y)=2x+2y+2; R is the triangle with corners (0,0), (1,0) and (0,1).

  3. 13.

    f(x,y)=x2+y2+10; R is the disk x2+y216.

  4. 14.

    f(x,y)=-2x+4y2+7 over R, the triangle bounded by y=-x, y=x, 0y1.

  5. 15.

    f(x,y)=23x3/2+2y3/2 over R, the rectangle with opposite corners (0,0) and (1,1).

  6. 16.

    f(x,y)=10-2x2+y2 over R, the disk x2+y225. (This is the cone with height 10 and base radius 5; be sure to compare your result with the known formula.)

  7. 17.

    Find the surface area of the sphere with radius 5 by doubling the surface area of f(x,y)=25-x2-y2 over R, the disk x2+y225. (Be sure to compare your result with the known formula.)

  8. 18.

    Find the surface area of the ellipse formed by restricting the plane f(x,y)=cx+dy+h to the region R, the disk x2+y21, where c, d and h are some constants. Your answer should be given in terms of c and d; why does the value of h not matter?

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