372 CHAPTER 20. THE INTEGRAL IN OTHER COORDINATES
of the form sin(φ)cos(θ)sin(φ)sin(θ)
cos(φ)
dρ
which points out from the surface of the sphere. Next keeping ρ and θ constant and differ-entiating only with respect to φ leads to an infinitesimal vector in the direction of a line oflongitude, ρ cos(φ)cos(θ)
ρ cos(φ)sin(θ)−ρ sin(φ)
dφ
and finally keeping ρ and φ constant and differentiating with respect to θ leads to the thirdinfinitesimal vector which points in the direction of a line of latitude. −ρ sin(φ)sin(θ)
ρ sin(φ)cos(θ)0
dθ
To find the increment of volume, we just need to take the absolute value of the determi-nant which has these vectors as columns, (Remember this is the absolute value of the boxproduct.) exactly as was the case for polar coordinates. This will also yield
dV = ρ2 sin(φ)dρdθdφ .
However, in contrast to the drawing of pictures, this procedure is completely generaland will handle all curvilinear coordinate systems and in any dimension. This is discussedmore later.
Example 20.3.3 Find the volume of a ball, BR of radius R. Then find∫
BRz2dV where z is
the rectangular z coordinate of a point.
In this case, U = (0,R]× [0,π]× [0,2π) and use spherical coordinates. Then this yieldsa set in R3 which clearly differs from the ball of radius R only by a set having volume equalto zero. It leaves out the point at the origin is all. Therefore, the volume of the ball is∫
BR
1dV =∫
Uρ
2 sinφ dV
=∫ R
0
∫π
0
∫ 2π
0ρ
2 sinφ dθ dφ dρ =43
R3π.
The reason this was effortless, is that the ball, BR is realized as a box in terms of thespherical coordinates. Remember what was pointed out earlier about setting up iteratedintegrals over boxes.
As for the integral, it is no harder to set up. You know from the transformation equationsthat z = ρ cosφ . Then you want∫
BR
zdV =∫ R
0
∫π
0
∫ 2π
0(ρ cos(φ))2
ρ2 sinφ dθ dφ dρ =
415
πR5
This will be pretty easy also although somewhat more messy because the function you areintegrating is not just 1 as it is when you find the volume.