Question

Integrate f(x,y,z)=12xz over the region in the first octant above the parabolic cylinder z=y^2 and below the paraboloid z=8−2x^2−y^2.

Answer 1

The intersection of the two surfaces occurs along the cylinder $x^2+y^2=4$, so we have the $x$- and $y$-coordinates of the points in the bounded region contained within $0\le x\le2$ and $0\le y\le2$, while $y^2\le z\le8-2x^2-y^2$.

The triple integral is then given by

$\displaystyle\iiint_V12xz\,\mathrm dV=\int_{x=0}^{x=2}\int_{y=0}^{y=2}\int_{z=y^2}^{z=8-2x^2-y^2}12xz\,\mathrm dz\,\mathrm dy\,\mathrm dx$

where $V$ denotes the bounded region in the first quadrant.

Integrating in Cartesian coordinates is easy enough to do in this order.

$\displaystyle\int_{x=0}^{x=2}\int_{y=0}^{y=2}\int_{z=y^2}^{z=8-2x^2-y^2}12xz\,\mathrm dz\,\mathrm dy\,\mathrm dx$
$=\displaystyle24\int_{x=0}^{x=2}\int_{y=0}^{y=2}x(x^2-4)(x^2+y^2-4)\,\mathrm dy\,\mathrm dx$
$=\displaystyle\int_{x=0}^{x=2}(512x-320x^3+48x^5)\,\mathrm dx$
$=256$