University level triple integration problem help.

Question

Use spherical coordinates to find the volume of the solid enclosed by the sphere $x^2+y^2+z^2=4a^2$ and the planes $z=0$ and $z=a$.

Answer 1

Below I've included a plot of the solid (in red) and its bounding surfaces. (Here $a=1$.)

One thing you can do is to find the volume of the cap (the region directly above the red solid bounded above by the sphere) and subtract that from the total volume of the upper half-sphere. In other words, if $D$ is the region whose volume you want to find and $D'$ is the cap region above $D$, then $$\iiint_D\mathrm{d}V=\frac{2}{3}\pi (2a)^3-\iiint_{D'}\mathrm{d}V$$ where $2a$ is obviously the radius of the sphere.

You have $$\iiint_{D'}\mathrm{d}V=\int_0^{2\pi}\int_0^{\pi/3}\int_{a\sec \phi}^{2a}\rho^2\sin\phi\,\mathrm{d}\rho\,\mathrm{d}\phi\,\mathrm{d}\theta$$ If you've done any integration in polar/cylindrical/spherical coordinates, then it should be clear why the azimuthal angle $0\le\theta\le2\pi$. Meanwhile, the polar angle $0\le\phi\le\dfrac{\pi}{3}$ because the intersection of the sphere with the plane occurs when $z=\rho\cos\phi=a$; substituting this into the sphere equation, you find that $\phi=\arctan\sqrt3=\dfrac{\pi}{3}$. Finally, since $a\le z\le \sqrt{4a^2-x^2-y^2}$, you have $a\sec s\le \rho\le2a$.