Question

Integrate f(x,y,z)=x over the region in the first octant (x>0, y>0, z>0) above z=y^2 and below z=8-2x^2-y^2

Answer 1

$\begin{cases}z=y^2\\z=8-2x^2-y^2\end{cases}\implies y^2=8-2x^2-y^2\iff y^2+x^2=4$

which means the intersection of the parabolic cylinder $z=y^2$ and paraboloid $z=8-2x^2-y^2$ is a circle of radius 2 centered at the origin.

So the integral can be represented in Cartesian coordinates by

$\displaystyle\iiint_D x\,\mathrm dV=\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{y^2}^{8-2x^2-y^2}x\,\mathrm dz\,\mathrm dy\,\mathrm dx$

where $D$ is the region between the two surfaces.

Converting to cylindrical coordinates will make this slightly easier to compute.

$\begin{cases}x(r,\theta,\zeta)=r\cos\theta\\y(r,\theta,\zeta)=r\sin\theta\\z(r,\theta,\zeta)=\zeta\end{cases}$
$\implies\dfrac{\partial(x,y,z)}{\partial(r,\theta,\zeta)}=\begin{vmatrix}x_r&x_\theta&x_\zeta\\y_r&y_\theta&y_\zeta\\z_r&z_\theta&z_\zeta\end{vmatrix}=\begin{vmatrix}\cos\theta&-r\sin\theta&0\\\sin t&r\cos t&0\\0&0&1\end{vmatrix}=r$

Letting $E$ denote the same region in cylindrical coordinates, you have

$\displaystyle\iiint_E r\cos\theta\left|\frac{\partial(x,y,z)}{\partial(r,\theta,\zeta)}\right|\,\mathrm dV=\int_0^{2\pi}\int_0^2\int_{r^2\sin^2\theta}^{8-r^2(1+\cos^2\theta)}r^2\cos\theta\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$

In either case the integral reduces to 0.