0
$\begingroup$

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

  • 0
    I like your idea of thinking of it as a solid of revolution. You can actually think of the caps of the tube as solids as well. Everything is symmetrical so find the cylinder's volume and find one cap's volume and you're done. You do have to find a point of intersection but that shouldn't be that difficult. – 2017-02-18

1 Answers 1

3

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;

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

  • 0
    Was looking for something like this, thank you! – 2017-02-18
  • 0
    @PeterGarder: You can accept this to close the problem. – 2017-03-04