# Need help with triple integral volume question

1. Dec 17, 2011

### mottov2

1. The problem statement, all variables and given/known data
A central cylinder of radius 1 is drilled out a sphere of radius 2. Let B be the region inside the sphere but outside the cylinder.
Evaluate

∫B 1/x2+y2+z2 dV

3. The attempt at a solution

Volume = ∫∫∫ 1/x2+y2+z2 dV = ∫∫∫ 1/$\rho$2 sin$\phi$$\rho$2 d$\rho$d$\theta$d$\phi$

where
0<$\phi$<pi
0<$\theta$<2pi
1<$\rho$<2

so i figured i should convert it to spherical coordinate and the final answer i get is 4pi.
Did i set up my integral properly? Is this question doable using cylindrical coordinate?

2. Dec 17, 2011

### Dick

The integral is ok. The bounds could use some work. I assume you putting the cylinder along the z-axis. phi doesn't go from 0 to pi and the inner value of r will depend on phi, won't it? Did you draw a sketch?

3. Dec 17, 2011

### mottov2

yes i put the cylinder along the z-axis.
ok so i drew projection on yz-axis which looks like two half circles with a gap in the middle.

so then would the phi ranges from 0 to pi/6, and rho range from 1/sin(phi) to 2?

which gives me an answer of 2pi(1-pi/6)

Last edited: Dec 17, 2011
4. Dec 18, 2011

### Dick

pi/6 is one value of phi where the cylinder intersects the cylinder alright. But so far your integral is over the top 'cap' of the sphere. You want the part outside of the cylinder. Check your sketch. But 1/sin(phi) looks good for a lower bound of r.

Last edited: Dec 18, 2011
5. Dec 18, 2011

### mottov2

Last edited by a moderator: May 5, 2017
6. Dec 18, 2011

### ehild

The sketch is all right, the range of phi is not. Phi is the angle the radius encloses with the positive z axis. It goes from pi/6, it is correct, but then it increases how far? to -phi/6?????
Think: without the cylinder, you would integrate from phi=0 to phi=-0?????

ehild

7. Dec 18, 2011

### mottov2

ohhhhhhh

ok so then phi should range from pi/6 to 5pi/6 ?!?!?!?!

8. Dec 18, 2011

### ehild

I think those are the correct bounds.

ehild