# Thread: integral (within sphere, below cone)

1. ## integral (within sphere, below cone)

Find the volume of the solid that lies within the sphere , above the plane, and outside the cone .

I am having trouble finding the integral of phi

p is from 0 to 1
theta is from 0 to 2pi
but the scale of 3 on the function is screwing me up with theta

i feel like im getting close, but i need a push
i thought it may have been 3*sqrt(1/10) to pi/2
but it didnt turn out to be

2. Well, we just want the volume of the cone with the sphere top then just subtract it out of the sphere right? Equation of the cone is $\displaystyle z=3\sqrt{x^2+y^2}$. So in spherical coordinates that's $\displaystyle \rho \cos(\phi)=3\sqrt{\rho^2\sin^2(\phi)}$. Do all that and get: $\displaystyle \rho^2\left(9\sin^2(\phi)-\cos^2(\phi)\right)=0$ or $\displaystyle \phi=\arctan(1/3)$. So equation of the cone in spherical coordinates is $\displaystyle \rho=\arctan(1/3)=a$. Do same with the sphere to get equation of it in spherical coordinates $\displaystyle \rho=1$. Anyway, turn the crank and get:

$\displaystyle V_{\text{cone}}=\int_0^a\int_0^{2\pi}\int_0^1 \rho^2\sin(\phi)d\rho d\theta d\phi$

Pretty sure anyway. Double check it ok.

3. im going to have a very busy day today, but ill make sure to check it out tomorrow for sure