Volume of $E=\{ (x,y,z)\in \mathbb R^3 | x^2+y^2\le4, x^2+y^2+z^2\le 16 \}$

Question

Compute the volume of $E=\{ (x,y,z)\in \mathbb R^3 | x^2+y^2\le4, x^2+y^2+z^2\le 16 \}$

I know it the inner tube of a ball but I can't use cylindrical coordinates because of the two curved ends, so I saw it as a surface of revolution and used coordinates

$r\in[0, \min(2,16-z^2))$

$\theta\in[0,2\pi)$

$ z\in[-4,4]$

But because of the minimum the computation will be a bit ugly and I was wondering if there is more elegant (or even standard) approach to this.

Thanks in advance

Answer 1

You domain of integration is given by $\{(\theta, r): \theta \in [0, 2\pi], r\in [0,2]\}$. Now parametrize the top of the sphere as $z = \sqrt{16-x^2-y^2} = \sqrt{16-r^2}$. Hence you have the volume is given by;

$$2\int_{0}^{2\pi} \int_{0}^2 r \int_{0}^{\sqrt{16-r^2}} 1\ dz dr d\theta$$