Question

Set up the integral needed to find the volume of the solid bounded by the hyperboloid z2 = 64 x2 y2 and by the upper nappe of the cone z2 = 2x2 2y2

Answer 1

$\begin{cases}z^2=64+x^2+y^2\\z^2=2x^2+2y^2\end{cases}$

It's clear enough that the upper half of the cone falls below the upper sheet of the paraboloid, so that $\sqrt{2x^2+2y^2}\le z\le\sqrt{64+x^2+y^2}$. Right away you can see that converting to cylindrical coordinates will be quite advantageous.

The intersection of the two surfaces occurs as a circle:

$64+x^2+y^2=2x^2+2y^2\implies64=x^2+y^2$

which is parallel to the x-y plane, has radius 8, and is centered at $(0,0,8\sqrt2)$.

The volume of this space is given by the integral

$\displaystyle\iiint_S\mathrm dV=\int_{x=-8}^{x=8}\int_{y=-\sqrt{64-x^2}}^{y=\sqrt{64-x^2}}\int_{z=\sqrt{2x^2+2y^2}}^{z=\sqrt{64+x^2+y^2}}\mathrm dz\,\mathrm dy\,\mathrm dz$

where $S$ denotes the bounded space between the surfaces. Converting to cylindrical coordinates, this can be expressed as

$\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=8}\int_{z=\sqrt2r}^{z=\sqrt{64+r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$

which evaluates to $\dfrac{1024\pi(\sqrt2-1)}3$.