Question

Find the volume of a solid enclosed by the paraboloid z = x2 +y2 and a plane z = 9

Answer 1

The plane $z=9$ lies above the paraboloid $z=x^2+y^2$, so the volume of the bounded region $R$ is given by

$\displaystyle\iiint_R\mathrm dV=\int_{-3}^3\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\int_{x^2+y^2}^9\mathrm dz\,\mathrm dy\,\mathrm dx$

Convert to cylindrical coordinates, setting

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=z\end{cases}\implies\mathrm dx\,\mathm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz$

and the integral is equivalent to

$\displaystyle\int_0^{2\pi}\int_0^3\int_{r^2}^9r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=2\pi\int_0^3(9r-r^3)\,\mathrm dr=\boxed{\frac{81\pi}2}$