27.3. CYLINDRICAL AND SPHERICAL COORDINATES 543

Example 27.3.6 Now remove the same two cones as in the above examples along with thesame slice and find the volume of what is left. Next, if R is the region just described, find∫

R xdV .

This time you need∫ 3π/4

π/6

∫ 2π

π/4

∫ R

0ρ

2 sin(φ)dρdθdφ =7

24

√2πR3 +

724

√3πR3

As to the integral, it equals∫ 3π/4

π/6

∫ 2π

π/4

∫ R

0(ρ sin(φ)cos(θ))ρ

2 sin(φ)dρdθdφ =− 1192

√2R4

(7π +3

√3+6

)This is because, in terms of spherical coordinates, x = ρ sin(φ)cos(θ).

Example 27.3.7 Set up the integrals to find the volume of the cone 0≤ z≤ 4,z=√

x2 + y2.Next, if R is the region just described, find

∫R zdV .

This is entirely the wrong coordinate system to use for this problem but it is a goodexercise. Here is a side view.

φ

You need to figure out what ρ is as a function of φ which goes from 0 to π/4. Youshould get ∫ 2π

0

∫π/4

0

∫ 4sec(φ)

0ρ

2 sin(φ)dρdφdθ =643

π

As to∫

R zdV, it equals

∫ 2π

0

∫π/4

0

∫ 4sec(φ)

0

z︷ ︸︸ ︷ρ cos(φ)ρ2 sin(φ)dρdφdθ = 64π

Example 27.3.8 Find the volume element for cylindrical coordinates.

In cylindrical coordinates,  xyz

=

 r cosθ

r sinθ

z

