# Homework Help: Double integral sphere/plane

1. Jul 19, 2011

### usfrbrpl

1. The problem statement, all variables and given/known data
Find the volume under the sphere x^2+y^2+z^2=r^2 and above the plane z=a, where 0<a<r

2. Relevant equations
x^2+y^2+z^2=r^2 is the equation of a sphere with radius r centered at the origin

z=a is the equation of a plane with height a parallel to the xy plane

V = ∫∫z dx dy over the region R

3. The attempt at a solution

I plugged z=a into the equation of the sphere x^2+y^2+z^2=r^2 to find where the two surfaces intersected. I got x^2+y^2=r^2-a^2. I also solved the original sphere equation for z to get z=(r^2-x^2-y^2)^.5, taking the positive root because the volume lies above the z axis.

V=∫∫zdxdy, I plugged in z=(r^2-x^2-y^2)^.5 to get ∫∫((r^2-x^2-y^2)^.5)dxdy

I solved for x in the intersection equation above to get my bounds of integration x=(r^2-a^2-y^2)^.5 and x=-(r^2-a^2-y^2)^.5

I set x=0 in the intersection equation above and solved for y to get my bounds of integration y=(r^2-a^2)^.5 and y=-(r^2-a^2)^.5

I put the above bounds next to their respective integral signs in the volume equation and put it in my calculator. It spit back a nasty looking formula at me, with sin(infinity) as part of the equation, so I think I did something wrong.

2. Jul 19, 2011

### LCKurtz

You want the volume above z = a so your integrand should be zupper- zlower so you need to subtract a from the integrand.

Next, I would use R instead of r as the radius of the sphere and change your dydx integral to polar coordinates. That way you won't confuse the radius of the sphere with the r in polar coordinates. It will simplify your integral a lot but may not make it that much easier to integrate.

3. Jul 20, 2011

### usfrbrpl

Thanks, I managed to get:

V = ∫∫r((R^2-r^2)^.5-a)drdθ with bounds of (R^2-r^2)^.5 to 0 on the dr and bounds of 2*pi to 0 on the dθ. Simplifying this gives me (2R^3-3a*R^2+a^3)*pi/3. Thanks again for your help!