\f752 Chapter 15. Curves, Surfaces, and Volumes Spherical coordinates: R. = pep dR = dpep + pape, + psin d ddeo p Isin d| do de (constant-p surface) (18) dA = p sin $| do de (constant-@ surface) pdp do (constant-A surface) dV = p' |sino] dodd de. where the absolute value signs are to be enforce that the area and volume elements always be positive. 15.6.2. Volume integrals. With dV in hand. as given by (4), we can deal with the volume integral ffff dV of a given function f over a given region ) in 3-space for if a, y, z are parametrizat 1, sav = / fatu, o, w) , y(u, u, w), =(up, w) ) O(x, y, =) O(u, v. w) du du dw, M . (19) where R is the region in u, v, w space corresponding to the region V in x, y, z space. We use single integral signs in ( 19) to denote triple integrals, for compactness, and Figure 4. Gravitational force use that notation routinely in Chapter 16. There should be no confusion, in (19), F exerted by M on m. because of the 1 and R. which are three-dimensional regions, and because the dV and du de do clearly denote triple integrals. EXAMPLE 3. Gravitational Attraction of a Cone. Newton's law of gravitation states that the force of attraction F exerted by any one point mass Af on any other point mass m is given by Mm F = G- - e, (20) where (Fig. 4) is the distance of separation. e is a unit vector directed from m toward A. and G (= 6.67 x 10-8 cm /g sec" ) is the universal gravitational constant; (20) is said to be an inverse square law since the force varies as the inverse square of the distance. (By A and m being point masses. we mean that their sizes are negligible compared with d.) If M is at (x, y, = ) and in is at (20; 90, 20), we can express e as 0 = (@ - en)i+ (y - yo)it (e - =)k (21 ) V(x - 20)' + (0 - 80)' + (2 - 20)? The problem that we pose here is to calculate the force of attraction exerted at the Figure 5. The cone. origin, per unit mass at the origin (he.. m = 1), by the solid right circular cone of uniform mass density o (g/em') shown in Fig. 5. Since the cone is not a point mass, we need to15.6. Volumes and Volume Integrals 753 find the forces induced by the individual mass elements that constitute the cone and to sum them by integration. With Af changed to o dV and do - yo = = = 0. (20) and (21) give F(0,0.0) = G JJJ aityitzk (3 + 2 + 23) 3/2 0 dV. (22) Surely, the symmetry about the = axis implies that the & and y components of F will turn out to be zero so that (22) can be simplified to F(0,0, 0) = Gok / / / (23) In what coordinate system shall we work out the integral? That is, how shall we parametriza the conical region? Spherical coordinates seem like a good choice inasmuch as the con- ical surface is a constant- coordinate surface. However, the top surface, = = h. is not a constant-coordinate surface. That is. p # constant, $ * constant, and 0 * constant on & = h. Using cylindrical coordinates the top surface is a constant-coordinate surface (namely, = = h) but the conical surface is not. Thus, the choice seems to be "six of one. half a dozen of the other." Let us work the problem both ways. Using cylindrical coordinates: That is, choose u, v, w to be the familiar cylindrical co- ordinates r, 0, ~, respectively. Then dV = r dr d0 dz and m' + y? = 7 so (23) becomes F=ock zr dr do dz ( 12 + =2)3/ 2 = 2AGGk ("=s' (1+tan' a) ; dude 213/2 (u = n'+ =" ) = 2ROCK /"= (- 30)| #* / cost a dz (24) so that F - 2noGh(1 - cosa)k. (25) In case the etan a limit was not clear, note that "/ = = tan a on the conical surface. Using spherical coordinates: This time choose u. , w to be the spherical coordinates p, d. 0, respectively. Then dV/ = p' sino dodo de, = = pcos , and a + 12 + 2 = p' so (23) becomes F = ock / JA. chi cos p cos (03jazz of sino do do de 777 / 1348 cos o sino- do = 2noGh(1 - cosa)k. (26) coSO754 Chapter 15. Curves, Surfaces, and Volumes which result agrees with (25). COMMENT. As a partial check, observe from (26) that F -+ 0 as a - 0, as it should. I Closure. With a three-dimensional region generated by a three-parameter family " = e(u,v, w), y = you, v, w), z = =(u, v, w), and the position vector given by R= = =(u, v, w)ity(u, v, w)j+z(u, v, w)k. the volume element dV is defined as the volume of the parallelipiped with edges Ru du, Ry du, Ry, dw as shown in Fig. 1. Thus, dV = ((Ru du) . (R. do) x (Ru du)) = [Ru . Ru X Ru| du du du yu Cu yu du du dw Du yw Ew du du dw. a (u. v, w) That is, dV is the absolute magnitude of the Jacobian of a, y, = with respect to u, v, w times du du dw. EXERCISES 15.6 2. Show by triple integration that the volume of the right cir- cular cone shown in Fig. 5 is V = (ah3/3) tance: (a) using cylindrical coordinates (b) using spherical coordinates