Question 2: Explain how the equation in this example is converted to spherical coordinates.

Hint 1: Read Example 4 (p. 946).

946 Chapter 15 Multiple Integrals The angle varies from 0 at the north pole of the sphere to /2 at the south pole; the angle # does not appear in the expression for p. reflecting the symmetry about the z-axis p = 2 cus (see Figure 15.53). EXAMPLE 4 Find a spherical coordinate equation for the cone z = Vr + y'. Solution 1 Use geometry. The cone is symmetric with respect to the z-axis and cuts the first quadrant of the yz-plane along the line z - y. The angle between the cone and the positive z-axis is therefore w/4 radians. The cone consists of the points whose spherical coordinates have o equal to w/4. so its equation is $ = 4/4. (See Figure 15.54.) Solution 2 Use algebra. If we use Equations (1) to substitute for x, y, and z we obtain the same result: FIGURE 15.53 The sphere in Example 3. : = Vr+> p cos 6 = Vp-sin' Example 3 p coso = p sin p 20, sind 2 0 cos = sind Spherical coordinates are useful for describing spheres centered at the origin, half- planes hinged along the z-axis, and cones whose vertices lie at the origin and whose axes lie along the z-axis. Surfaces like these have equations of constant coordinate value: P=4 Sphere, radius 4, center at origin Cone opening up from the origin, making an angle of w/3 radians with the positive 2-axis Half-plane, hinged along the z-axis, making an FIGURE 15.54 The cone in Example 4. angle of w/3 radians with the positive .x-axisWhen computing triple integrals over a region D in spherical coordinates, we partition the region into n spherical wedges. The size of the kth spherical wedge, which contains a P sin h point (p.. 4, 0), is given by the changes Ap., Ads, and AN, in p. 6, and 0. Such a p sin An spherical wedge has one edge a circular arc of length p, Ad, another edge a circular arc of length p, sin d: 46, and thickness App. The spherical wedge closely approximates a cube of these dimensions when App, Ad, and Ad, are all small (Figure 15.55). It can be shown that the volume of this spherical wedge AV is AV = p, sind, Ap, Ad, A6, for (pe, 8), a point chosen inside the wedge. The corresponding Riemann sum for a function f(p, 5, 0) is S, = > f(p. 4, 03 pi sin d, Ap, Ad, 40,. Ap As the norm of a partition approaches zero, and the spherical wedges get smaller, the Riemann sums have a limit when f is continuous: FIGURE 15.65 In spherical coordinates we use the volume of a spherical wedge, which closely approximates that of a cube. lim S, = f(p. b, 0) dV = / f( p. do, @) ph sin do dp led do. Volume Differential in Spherical Coordinates To evaluate integrals in spherical coordinates, we usually integrate first with respect dV = p sin o dp do do to p. The procedure for finding the limits of integration is as follows. We restrict our