Originally Posted by

**arbolis** I didn't reach the result yet and I probably made an error somewhere.

Calculus of the volume of the cylinder inside the sphere : $\displaystyle I=2\int_{0}^{\frac{\pi}{2}} \int_{0}^{2a\cos (\theta)} \int_{-\sqrt{4a^2-r^2}}^{\sqrt{4a^2-r^2}} rdzdrd\theta$.

$\displaystyle \int_{-\sqrt{4a^2-r^2}}^{\sqrt{4a^2-r^2}} rdz=2r\sqrt{4a^2-r^2}$.

So $\displaystyle I=2\int_{0}^{\frac{\pi}{2}} \int_{0}^{2a\cos (\theta)} 2r\sqrt{4a^2-r^2} dr d\theta$.

I call $\displaystyle J=\int_{0}^{2a\cos(\theta)} 2r\sqrt{4a^2-r^2}dr$.

Let $\displaystyle u=4a^2-r^2 \Rightarrow du=-2rdr$, hence $\displaystyle {\color{red}J=-\int_{0}^{2a\cos (theta)} \sqrt{u} du=-\frac{2}{3} (2a\cos(\theta))^{\frac{3}{2}}}$.

Thus $\displaystyle I=-\frac{4}{3} \int_{0}^{\frac{\pi}{2}} (2a\cos (\theta))^{\frac{3}{2}}d\theta$.

I don't know how to solve this integral. I didn't reach $\displaystyle \sin ^3 (\theta)$ yet so I think I made an error.

Could you tell me if I made an error? If no, I'd like to know which method to use to solve the integral because I've no idea.

Thank you for your help.