# Homework Help: Find the volume: once more

1. Jul 16, 2011

### stratusfactio

1. The problem statement, all variables and given/known data
The textbook I have is horrible in explaining this stuff in the chapter so please bare with me.

So we have the information: Find the volume V inside both the sphere x^2 + y^2 + z^2 = 1 and the cone z =sqrt(x^2 + y^2)

2. Relevant equations
Where the heck do I began? I know this will require triple integration in spherical coordinates and the solution is (2/3)*pi*1^3(1-1/sqrt(2))

3. The attempt at a solution
I know the theta integrand is (0,2pi) because there are no bounds, but I have no idea where to find the limits of integration for the phi and zenith integrands. :(

EDIT (1): Just solved for the intersections of the sphere and cone and got z = 1/sqrt(2))
EDIT (2): Since phi = 1 and z = (1/sqrt(2)), it's safe to conclude that the zenith angle is pi/4...but what would be it's upper limit?
EDIT (3): I believe the limits for phi is 0 to pi/4. But I'm not quite sure why though? I think I got the integrands all figured out. I believe it's: for p = (0,1), phi = (0, pi/4) and theta = (0,2pi).

So overall, could someone just briefly explain to me why phi's limits of integration is from 0 to pi/4. thanks a bunch :D

Last edited: Jul 16, 2011
2. Jul 17, 2011

### HallsofIvy

Looking at the figure from the side (along the negative y-axis), you should see a circle and a "V" (y= 0 so the equations are $x^2+ z^2= 1$, a circle, and $z= \sqrt{x^2}= |x|$). The "co-latitude", $\phi$, goes from directly up the z-axis, $\phi= 0$, to the line z= x, which bisects the right angle between the x and z axes and so makes angle $\pi/4$ with the z-axis.

3. Jul 17, 2011

### LCKurtz

I'm sorry. Undressing is forbidden in this forum.