What is the increment of volume in spherical coordinates? There are two ways to see what this is, through art and through a systematic procedure. First consider art. Here is a picture.
In the picture there are two concentric spheres formed by making ρ two different constants and surfaces which correspond to θ assuming two different constants and ϕ assuming two different constants. These intersecting surfaces form the little box in the picture. Here is a more detailed blow up of the little box.
What is the volume of this little box? Length ≈ ρdϕ, width ≈ ρsin

Now what is really going on? Consider the dot in the picture of the little box. Fixing θ and ϕ at their values at this point and differentiating with respect to ρ leads to a little vector of the form

which points out from the surface of the sphere. Next keeping ρ and θ constant and differentiating only with respect to ϕ leads to an infinitesimal vector in the direction of a line of longitude,

and finally keeping ρ and ϕ constant and differentiating with respect to θ leads to the third infinitesimal vector which points in the direction of a line of latitude.

To find the increment of volume, we just need to take the absolute value of the determinant which has these vectors as columns, (Remember this is the absolute value of the box product.) exactly as was the case for polar coordinates. This will also yield

However, in contrast to the drawing of pictures, this procedure is completely general and will handle all curvilinear coordinate systems and in any dimension. This is discussed more later.
Example 15.3.3 Find the volume of a ball, B_{R} of radius R. Then find ∫ _{BR}z^{2}dV where z is the rectangular z coordinate of a point.
In this case, U = (0,R] ×
As for the integral, it is no harder to set up. You know from the transformation equations that z = ρcosϕ. Then you want

This will be pretty easy also although somewhat more messy because the function you are integrating is not just 1 as it is when you find the volume.
Example 15.3.4 A cone is cut out of a ball of radius R as shown in the following picture, the diagram on the left being a side view. The angle of the cone is π∕3. Find the volume of what is left.
Use spherical coordinates. This volume is then

Now change the example a little by cutting out a cone at the bottom which has an angle of π∕2 as shown. What is the volume of what is left?
This time you would have the volume equals

Example 15.3.5 Next suppose the ball of radius R is a sort of an orange and you remove a slice as shown in the picture. What is the volume of what is left? Assume the slice is formed by the two half planes θ = 0 and θ = π∕4.
Using spherical coordinates, this gives for the volume

Example 15.3.6 Now remove the same two cones as in the above examples along with the same slice and find the volume of what is left. Next, if R is the region just described, find ∫ _{R}xdV .
This time you need

As to the integral, it equals

This is because, in terms of spherical coordinates, x = ρsin
Example 15.3.7 Set up the integrals to find the volume of the cone 0 ≤ z ≤ 4,z =
This is entirely the wrong coordinate system to use for this problem but it is a good exercise. Here is a side view.
You need to figure out what ρ is as a function of ϕ which goes from 0 to π∕4. You should get

As to ∫ _{R}zdV, it equals

Example 15.3.8 Find the volume element for cylindrical coordinates.
In cylindrical coordinates,

Therefore, the Jacobian determinant is

It follows the volume element in cylindrical coordinates is rdθdrdz.
Example 15.3.9 In the cone of Example 15.3.7 set up the integrals for finding the volume in cylindrical coordinates.
This is a better coordinate system for this example than spherical coordinates. This time you should get

Example 15.3.10 This example uses spherical coordinates to verify an important conclusion about gravitational force. Let the hollow sphere, H be defined by a^{2} < x^{2} + y^{2} + z^{2} < b^{2}
and suppose this hollow sphere has constant density taken to equal 1. Now place a unit mass at the point
Without loss of generality, assume z_{0} > 0. Let dV be a little chunk of material located at the point


Therefore, the total force is

By the symmetry of the sphere, the i and j components will cancel out when the integral is taken. This is because there is the same amount of stuff for negative x and y as there is for positive x and y. Hence what remains is

as claimed. Now for the interesting part, the integral is evaluated. In spherical coordinates this integral is.
 (15.1) 
Rewrite the inside integral and use integration by parts to obtain this inside integral equals


 (15.2) 
There are some cases to consider here.
First suppose z_{0} < a so the point is on the inside of the hollow sphere and it is always the case that ρ > z_{0}. Then in this case, the two first terms reduce to

and so the expression in 15.2 equals




Therefore, in this case the inner integral of 15.1 equals zero and so the original integral will also be zero.
The other case is when z_{0} > b and so it is always the case that z_{0} > ρ. In this case the first two terms of 15.2 are

Therefore in this case, 15.2 equals

and so