Volume using triple integrals

[PhysDrew]

Homework Statement

Use a triple integral to calculate the volume of the solid enclosed by the sphere
x^2 + y^2 + z^2=4a^2 and the planes z=0 and z=a

Homework Equations

Transform to spherical coordinates (including the Jacobian)

The Attempt at a Solution

I'm stuck, as the radius of the sphere is not constant through the area of integration due to the plane z=a. It looks like I should split this integral up, but I'm just not sure on how to do it. It looks like the angle rho is (pi)/3 when the radius of the sphere(2a, from the origin) hits the z=a plane. Please help!

 

[HallsofIvy]

Yes, for $\phi< \pi/3$ the upper boundary is the z= a plane. So, going up from the origin to z= a, then over to the angle $\phi$, you have a right triangle with angle $\phi$ and near side of length a. $\rho$ is the length of the hypotenuse of that right triangle. $\cos(\phi)= a/\rho$ so
$$\rho= \frac{a}{\cos(\phi)}= a \sec(\phi)$$.

 

[PhysDrew]

Yes, for $\phi< \pi/3$ the upper boundary is the z= a plane. So, going up from the origin to z= a, then over to the angle $\phi$, you have a right triangle with angle $\phi$ and near side of length a. $\rho$ is the length of the hypotenuse of that right triangle. $\cos(\phi)= a/\rho$ so
$$\rho= \frac{a}{\cos(\phi)}= a \sec(\phi)$$.

Oh good so I was right, thanks. So should I integrate with rho being between 0 and 2a, with phi being between pi/2 and pi/3, and then...well I get stuck there. I've got that little cone bit in the middle to go and I don't know how to get him (or her). Sorry I'm just not getting this one