I have the find the volume of the region bounded by the paraboloid $y = 2x^2 +2z^2$ and the plane $y=8$. Is the volume (using triple integrals) just $$\int_{-2}^2\int_{-2}^{2}\int_{2x^2+2z^2}^8dydzdx$$
I understand the bounds for y (from the paraboloid to the plane), but I don't understand what are the bounds for x and z.
[Math] Volume of the Region bounded by $y = 2x^2 +2z^2$ and the plane $y=8$
3dintegrationmultivariable-calculusvolume
Best Answer
$y = 2x^2 +2z^2$ and the plane $y=8$ gives:
$8=2x^2 +2z^2$
$4=x^2 +z^2$
This is a circle and gives you limits for $x$ and $z$ obviously it will be polar coordinates problems in xz space.
Limits for y are solved because your surface is going from paraboloid to the $y=8$, so modify your integral bounds... to be something like: $x=rcos(\phi), z=rsin(\phi)$, $y=y$:
$$\int_{0}^{2\pi}d\phi\int_{0}^{2}dr\int_{2r^2}^8 rdy$$
projecting the section of paraboloid and $y=8$ you get xz projection which defines you x and z bounds.