Volume of the bounded by the region $z=9-x^2-y^2$ and the plane $z=0$

Question

I have to find the volume bounded by the equations $z=9-x^2-y^2$ and $z=0$, but I do not know how it is supposed to be done correctly. I have right now:

$\int_0^{2\pi}\int_0^3 (9-r^2)rdrd\phi = \frac{243}{2}\pi$

But I do not know if this is correct or if it is supposed to be done in this way. Thanks for the help.

Answer 1

In cylindrical coordinates, the computation of that volume becomes$$\int_0^{2\pi}\int_0^9\int_0^{\sqrt{9-z}}r\,\mathrm dr\,\mathrm dz\,\mathrm d\theta=\frac{81\pi}2.$$