Question

Triple integral where e is enclosed by paraboloid z= x^2 y^2 and plane z=4

Answer 1

Without any more information, I can only assume you're trying to find the volume of this bounded region. In that case, you're computing

$\displaystyle\iiint_E\mathrm dV=\iiint_E1\,\mathrm dx\,\mathrm dy\,\mathrm dz$

This can be written as

$\displaystyle\int_{x=-2}^{x=2}\int_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}}\int_{z=x^2+y^2}^{z=4}\mathrm dz\,\mathrm dy\,\mathrm dx$

Converting to cylindrical coordinates will make this easier.

$\begin{cases}\mathbf x(r,\theta,z)=r\cos\theta\\\mathbf y(r,\theta,z)=r\sin\theta\\\mathbf z(r,\theta,z)=z\end{cases}\implies\dfrac{\partial(\mathbf x,\mathbf y,\mathbf z)}{\partial(r,\theta,z)}=\left|\det\begin{bmatrix}\mathbf x_r&\mathbf y_r&\mathbf z_r\\\mathbf x_\theta&\mathbf y_\theta&\mathbf z_\theta\\\mathbf x_z&\mathbf y_z&\mathbf z_z\end{bmatrix}\right|=r$

Now, in cylindrical coordinates the integral is given by

$\displaystyle\iiint_E\mathrm dV=\iiint_Er\,\mathrm dr\,\mathrm d\theta\,\mathrm dz$
$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=2}\int_{z=r^2}^{z=4}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=8\pi$