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`