How do I used triple integrals to calculate the volume of this solid?

Question

The questions reads:

Evaluate the iterated integral in spherical coordinates that describes the volume of the solid enclosed by the sphere $x^2 + y^2 + z^2 = 4a^2$ and the planes $z = 0$ and $z = a$.

I understand how to convert this to polar coordinates, but when I look at the answer the book provides, I don't understand why they split it into the sum of 2 triple integrals. I also don't understand why the bound on the phi integral in the second triple integral goes from $\pi/3$ to $\pi/2$.

This is the answer in the book:

Answer 1

Her is a side-long view of the figure.

One integral give you the volume of the green region where $\rho = 2a$ and one give you the volume of the red region where $\rho = a\sec\phi$

Answer 2

Essentially the split has to do with the fact that it is convenient to consider how the integral behaves in two different regions.

The first region is the one where the upper plane is outside the sphere. In this region the integral is just like that of a normal sphere, since the plane is outside of it and doesn't 'slice any volume off'.
This region corresponds to elevation angles between $0$ and $\frac{\pi}{3}$.

The second region is the one where the upper plane is inside the sphere. In this region the volume of the sphere is 'cut off' by the plane, and so the radial path over which you integrate is shorter here. $a\space sec\phi$ instead of $2a$.
This region corresponds to elevation angles between $\frac{\pi}{3}$ and $\frac{\pi}{2}$.

Answer 3

The intersection between the sphere and the plane $z=0$ leaves you the north hemisphere this implies the volume enclosed is $$V_{1}=\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{0}^{\rho\sec\phi}\! \rho^{2}\sin\phi \,d\rho\,d\phi\,d\theta$$ because the radius of the sphere is $\rho=2a$ and we know $\rho=\frac{z}{\cos\phi}=z\sec\phi$. So for the intersection with the plane $z=a$, we have the relation $z=\rho\cos\phi \Rightarrow \cos\phi=\frac{a}{2a}=\frac{1}{2} \Rightarrow \phi=\frac{\pi}{3}$, this implies and the volume enclosed is $$V_{2}=\int_{0}^{2\pi}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\int_{0}^{2a}\! \rho^{2}\sin\phi \,d\rho\,d\phi\,d\theta$$ Finally, the total volume is the sum of two integrals.