1. Multiple Integrals 12

2. Surface Area 12.6

3. 3 Surface Area In this section we apply double integrals to the problem of computing the area of a surface. We start by finding a formula for the area of a parametric surface and then, as a special case, we deduce a formula for the surface area of the graph of a function of two variables.

4. 4 Surface Area Recall that a parametric surface S is defined by a vector-valued function of two parameters r(u, v) = x(u, v) i + y(u, v) j + z(u, v) k or, equivalently, by parametric equations x = x(u, v) y = y(u, v) z = z(u, v) where (u, v) varies throughout a region D in the uv-plane. We will find the area of S by dividing S into patches and approximating the area of each patch by the area of a piece of a tangent plane. So first let’s recall how to find tangent planes to parametric surfaces.

5. 5 Surface Area Let P 0 be a point on S with position vector r(u 0, v 0 ). If we keep u constant by putting u = u 0, then r(u 0, v) becomes a vector function of the single parameter v and defines a grid curve C 1 lying on S. (See Figure 1.) Figure 1

6. 6 Surface Area The tangent vector to C 1 at P 0 is obtained by taking the partial derivative of r with respect to v: Similarly, if we keep v constant by putting v = v 0, we get a grid curve C 2 given by r(u, v 0 ) that lies on S, and its tangent vector at P 0 is

7. 7 Surface Area If the normal vector r u  r v is not 0, then the surface S is called smooth. (It has no “corners”.) In this case the tangent plane to S at P 0 exists and can be found using the normal vector. Now we define the surface area of a general parametric surface given by Equation 1. For simplicity we start by considering a surface whose parameter domain D is a rectangle, and we divide it into subrectangles R ij.

8. 8 Surface Area Let’s choose to be the lower left corner of R ij. (See Figure 2.) Figure 2 The image of the subrectangle R ij is the patch S ij.

9. 9 The part S ij of the surface S that corresponds to R ij is called a patch and has the point P ij with position vector as one of its corners. Let be the tangent vectors at P ij as given by Equations 3 and 2. Surface Area

10. 10 Surface Area Figure 3(a) shows how the two edges of the patch that meet at P ij can be approximated by vectors. These vectors, in turn, can be approximated by the vectorsand because partial derivatives can be approximated by difference quotients. So we approximate S ij by the parallelogram determined by the vectors and. Figure 3(a) Approximating a patch by a parallelogram

11. 11 Surface Area As shown in Figure 3(b), this parallelogram lies in the tangent plane to S at P ij. The area of this parallelogram is and so an approximation to the area of S is Figure 3(b) Approximating a patch by a parallelogram

12. 12 Surface Area Our intuition tells us that this approximation gets better as we increase the number of subrectangles, and we recognize the double sum as a Riemann sum for the double integral  D | r u  r v | du dv. This motivates the following definition.

13. 13 Example 1 – Area of a Parametric Surface Find the surface area of a sphere of radius a. Solution: We have found the parametric representation x = a sin  cos  y = a sin  sin  z = a cos  where the parameter domain is D = {( ,  ) | 0    , 0    2  } We first compute the cross product of the tangent vectors:

14. 14 Example 1 – Solution Thus since sin   0 for 0    . cont’d

15. 15 Example 1 – Solution Therefore, by Definition 4, the area of the sphere is cont’d

16. 16 Surface Area of a Graph

17. 17 Surface Area of a Graph For the special case of a surface S with equation z = f (x, y), where (x, y) lies in D and f has continuous partial derivatives, we take x and y as parameters. The parametric equations are x = x y = y z = f (x, y) so and

18. 18 Surface Area of a Graph Thus the surface area formula in Definition 4 becomes

19. 19 Example 2 – Surface Area of a Graph Find the area of the part of the paraboloid z = x 2 + y 2 that lies under the plane z = 9. Solution: The plane intersects the paraboloid in the circle x 2 + y 2 = 9, z = 9. Therefore the given surface lies above the disk D with center the origin and radius 3. (See Figure 4.) Figure 4

20. 20 Example 2 – Solution Using Formula 6, we have cont’d

21. 21 Example 2 – Solution Converting to polar coordinates, we obtain cont’d

22. 22 Surface Area of a Graph A common type of surface is a surface of revolution S obtained by rotating the curve y = f (x), a  x  b, about the x-axis, where f (x)  0 and f is continuous.