Need help with triple integral volume question

[mottov2]

Homework Statement

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

The Attempt at a Solution

Volume = ∫∫∫ 1/x²+y²+z² dV = ∫∫∫ 1/\rho² sin\phi\rho² d\rhod\thetad\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?

 

[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?

 

[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)

 

[Dick]

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)

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.

 

[mottov2]

ok then the phi would range from -pi/6 to pi/6?
this is the picture of the solid
http://img444.imageshack.us/img444/4821/sphereg.jpg and the picture i drew on the yz axis
http://img522.imageshack.us/img522/9702/tripinteg2.jpg did i sketch this out properly?

 

[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

 

[mottov2]

ohhhhhhh

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

 

[ehild]

ohhhhhhh

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

I think those are the correct bounds.

ehild