SURFACE INTEGRALS

5. SURFACE INTEGRALS

In the preceding sections we found surface areas, moments of them, etc., for surfaces of revolution. We now want to consider a way of computing surface integrals in general whether the surface is a surface of revolution or not. Consider a part of a surface as in Figure 5.1 and its projection in the (x, y) plane. We assume that any line parallel to the z axis intersects the surface only once. If this is not true, we must work with part of the surface at a time, or project the surface into a different plane. For example, if the surface is closed, we could find the areas of the upper and lower parts separately. For a cylinder with axis parallel to the z axis we could project the front and back parts separately into the (y, z) plane.

Figure 5.1

Let dA (Figure 5.1) be an element of surface area which projects onto dx dy in the (x, y) plane and let γ be the acute angle between dA (that is, the tangent plane at dA) and the (x, y) plane. Then we have

(5.1) dx dy = dA cos γ

or

dA = sec γ dx dy.

Section 5 Surface Integrals 271

The surface area is then (5.2)

dA =

sec γ dx dy

where the limits on x and y must be such that we integrate over the projected area in the (x, y) plane.

Now we must find sec γ. The (acute) angle between two planes is the same as the (acute) angle between the normals to the planes. If n is a unit vector normal to the surface at dA (Figure 5.1), then γ is the (acute) angle between n and the z axis, that is, between the vectors n and k, so cos γ = |n · k|. Let the equation of the surface be φ(x, y, z) = const. Recall from Chapter 4 just after equation (9.14) that the vector

grad φ = i

is normal to the surface φ(x, y, z) = const. (Also see Chapter 6, Section 6.) Then n is a unit vector in the direction of grad φ, so

n = (grad φ)/| grad φ|.

From (5.3) and (5.4) we find

k · grad φ

Often the equation of a surface is given in the form z = f (x, y). In this case φ(x, y, z) = z − f(x, y), so ∂φ/∂z = 1, and (5.5) simplifies to

sec γ = (∂f/∂x) 2 + (∂f /∂y) 2 + 1.

We then substitute (5.5) or (5.6) into (5.2) and integrate to find the area. To find centroids, moments of inertia, etc., we insert the proper factor into (5.2) as we have discussed in Section 3.

272 Multiple Integrals; Applications of Integration Chapter 5

Example 1. Find the area cut from the upper half of the sphere

x 2 +y 2 +z 2 = 1 by the cylinder x 2 +y 2 − y = 0.

This is the same as the area on the sphere which projects onto the disk x 2 +y 2 − y ≤ 0 in the (x, y) plane. Thus we want to integrate (5.2) over the area of this disk. Figure

5.2 shows the disk of integration (shaded) and the equatorial circle of the sphere (large circle). We compute sec γ from the equation of the sphere; we could use (5.6), but it is easier in

Figure 5.2 this problem to use (5.5):

φ=x 2 +y 2 +z 2 ,

| grad φ|

(2x) 2 + (2y) 2 + (2z) 2 1 = 1 = |∂φ/∂z|

sec γ =

. We find the limits of integration from the equation of the shaded disk, x 2 +y 2 −y ≤ 0.

2z

1 − x 2 −y 2

Because of the symmetry we can integrate over the first-quadrant part of the shaded area and double our result. Then the limits are:

x from 0 to y − y 2 , y from 0 to 1.

The desired area is

This integral is simpler in polar coordinates. The equation of the cylinder is then r = sin θ, so the limits are: r from 0 to sin θ, and θ from 0 to π/2. Thus (5.7) becomes

√ This is still simpler if we make the change of variable z = 1−r 2 √ . Then dz =

−r dr/ 1−r 2 , and the limits r = 0 to sin θ become z = 1 to cos θ. Thus (5.8) becomes

PROBLEMS, SECTION 5

For these problems, the most important sketch is the projection in the plane of integration, which is easy to do by hand. However, you might like to use your computer to plot the corresponding 3 dimensional picture.

1. Find the area of the plane x − 2y + 5z = 13 cut out by the cylinder x 2 +y 2 = 9. 2. Find the surface area cut from the cone 2x 2 + 2y 2 = 5z 2 , z > 0, by the cylinder

x 2 +y 2 = 2y. 3. Find the area of the paraboloid x 2 +y 2 = z inside the cylinder x 2 +y 2 = 9.

Section 6 Miscellaneous Problems 273

4. Find the area of the part of the cone 2z 2 =x 2 +y √ 2 in the first octant cut out by the planes y = 0, and y = x/

3, and the cylinder x 2 +y 2 = 4. 5. Find the area of the part of the cone z 2 = 3(x 2 +y 2 ) which is inside the sphere

x 2 +y 2 +z 2 = 16. 6. In Example 1, find the area of the cylinder inside the sphere. 7. Find the area of the part of the cylinder y 2 +z 2 = 4 in the first octant, cut out by

the planes x = 0 and y = x. 8. Find the area of the part of the cylinder z = x + y 2 that lies below the second-

quadrant area bounded by the x axis, x = −1, and y 2 = −x. 9. Find the area of the part of the cone x 2 +y 2 =z 2 that is over the disk (x−1) 2 +y 2 ≤ 1.

10. Find the area of the part of the sphere of radius a and center at the origin which is √ √ above the square in the (x, y) plane bounded by x = ±a/

2 and y = ±a/ 2. Hint for evaluating the integral : Change to polar coordinates and evaluate the r integral first.

11. The part of the plane x + y + z = 1 which is in the first octant is a triangular area (sketch it). Find the area and its centroid by integration. You might like to check your work by geometry.

12. In Problem 11, let the triangle have a density (mass per unit area) equal to x. Find the total mass and the coordinates of the center of mass.

13. For the area of Example 1, find the z coordinate of the centroid. 14. For the area in Example 1, let the mass per unit area be equal to |x|. Find the total

mass. 15. For a uniform mass distribution over the area of Example 1, find the moment of

inertia about the z axis. 16. Find the centroid of the surface area in Problem 2.