# Volume using triple integrals

## 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!

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
$$\rho= \frac{a}{\cos(\phi)}= a \sec(\phi)$$.
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
$$\rho= \frac{a}{\cos(\phi)}= a \sec(\phi)$$.