# Triple Intergral Problem

1. Jun 8, 2007

### FunkyDwarf

Hey guys!

Just a quick question. As usual im sure im missing something stupid but if you can help me get my head around it id appriciate it (ie sorry if it seems a mundane error)

1. The problem statement, all variables and given/known data
Find the mass of the solid T outside the sphere $$x^2 + y^2 + z^2 = 1$$
and inside the sphere $$x^2 +y^2 + z^2 = 2z$$

2. Relevant equations
Ok clearly given theyre spheres spherical coordinates is the go, but i kinda hit a stang finding the boundary for the radius or rho.

3. The attempt at a solution
Clearly the first sphere has radius one, but what about the 2nd one? Given its related to 2z it would seem more of an ellipse type thing...

given that the LHS is r^2 i used the relation z = rcos(phi) which gives the upper bound of r as cos(phi)

Does that sound right?

Cheers
-G

2. Jun 8, 2007

### Office_Shredder

Staff Emeritus
$$x^2+y^2+z^2=2z$$ isn't a sphere, so don't try and make it one.

you might want to start by noting if z<1/2, the x-y cross section of the outer shell is smaller than the x-y cross section of the inner shell... this gives you one bound for z. Try and find the other one.

3. Jun 8, 2007

### FunkyDwarf

Ah ok fair nuff, will do. Yeh i thought it wasnt a sphere, but the question stated it as one so guess its a misprint. thanks, will give it a go.

EDIT: Ok tried drawing some level curves...not sure theyre right. am i correct in thinking the 2nd function (2z) looks something like a y=x^2 curve but rotated around the y axis (centered about z here though)?

if so how can we use that as our upper bound because it is unbounded (in the z variable anyway) ...

sorry if seems like im making stupid oversights :S

cheers
-G

Last edited: Jun 8, 2007
4. Jun 8, 2007

### NateTG

$$x^2+y^2+z^2=2z$$
$$x^2+y^2+z^2-2z=0$$
$$x^2+y^2+z^2-2z+1=1$$
$$x^2+y^2+\left(z-1\right)^2=1$$

Looks a whole lot like a sphere of radius 1, centered at $<0,0,1>$ to me.

This looks very much like a question you're supposed to answer by setting up an integral.

5. Jun 8, 2007

### FunkyDwarf

Knew it was something simple :) guess i shoulda done a few more iterations of th level surface huh

thanks mate
-G