# Homework Help: Finding the volume using spherical coordinates

1. Apr 5, 2014

### Temp0

1. The problem statement, all variables and given/known data

Let V be the volume of the solid enclosed by the sphere x^2 + y^2 + z^2 - 2z = 0 ,
and the hemisphere x^2 + y^2 + z^2 = 9 , z ≥ 0. Find V

2. Relevant equations
Using spherical coordinates:
x^2 + y^2 + z^2 = ρ^2
z = ρcos(ø)

3. The attempt at a solution
So I changed both of them to spherical coordinates, which I get ρ = 3 and ρ = 2 cos (ø). I then attempt to use triple integration, solving ∫∫∫dV where dV = ρ^2 sin(ø) dρdødθ
I find that the domain for θ is [0 , 2∏], and the domain for ø is [0, ∏/2]. However, I'm having trouble finding the domain for ρ. From where to where do I integrate? Any help would be appreciated, thank you in advance.

2. Apr 5, 2014

### pasmith

There are two better ways to do this than using spherical polars:
(1) Work out the radius of the sphere $x^2 + y^2 + z^2 - 2z = 0$ and subtract its volume from that of the hemisphere.
(2) Use cylindrical polar coordinates.

3. Apr 5, 2014

### HallsofIvy

The first thing I would do is add 1 to both sides of $x^2+ y^2+ z^2- 2z= 0$ to get $x^2+ y^2+ z^2- 2z+ 1= x^2+ y^2+ (z- 1)^2= 1$ which says that the sphere has center at (0, 0, 1) and radius 1. That is completely contained in the upper hemisphere of a sphere centered at (0, 0, 0) and radius 3. That is why pasmith says you can just find the volume of the hemisphere and subtract from it the volume of the sphere.

4. Apr 5, 2014

### Temp0

I did it with the method where you subtract the larger volume from the smaller one to find the volume in between, but I want to check with the cylindrical method. I'm having some trouble with it because if I isolate z, then it becomes a huge mess of things and seems impossible to solve.

5. Apr 5, 2014

### LCKurtz

You don't want cylindrical coordinates for this problem if you are going to do the integrals. You want to write the smaller sphere $x^2+y^2+z^2-2z =0$ directly in spherical coordinates$$\rho^2 - 2\rho\cos\phi = 0$$so you have $\rho=2\cos\phi$ for the inner surface and $\rho=3$ for the outer surface. Set it up in spherical coordinates.

 I see now that the $\rho$ limits answer your original question.