I tried to solve it by spliting the volume in two, by the $xy$-axis:
$$1/2 V = \iiint r\,\mathrm{d}z\,\mathrm{d}r\,\mathrm{d}\theta$$
where the integral is bounded by:
- $0 < z < \sqrt{4-r^2}$
- $0 < r < 2\sin(\theta)$
- $0 < \theta < \pi$
This is equal to $1/2 V = 8\pi/3$. So the whole wolume is:
$$V = 16\pi/3.$$
Is this correct? Or did I do make some wrong presumptions about the boundary of the integral?
One more question: When I used online calculators I either get $16\pi/3$ or $8/9(3\pi - 4)$, why is that?
The limits are fine but the result is not.
As you may realize that the total volume of the sphere is $\displaystyle \frac{32\pi}{3}$ and volume of half sphere for $y \geq 0$ is $\displaystyle \frac{16\pi}{3}$. The cylinder only intersects the sphere in the region $y \geq 0$ and the intersected volume is not clearly the whole volume of the half sphere. So logically $\displaystyle \frac{16\pi}{3}$ does not seem the right answer.
As there is symmetry with respect to $YZ$ plane, my suggestion is to use the integral with bounds $0 \leq \theta \leq \frac{\pi}{2}$.
So, $V = \displaystyle 2 \int_0^{\pi/2} \int_0^{2 \sin\theta} \int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} r \ dz \ dr \ d\theta = \frac{16 \pi}{3} - \frac{64}{9}$.