Question

Integrate f(xyz)=8xz 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 surfaces is

\[8-2x ^{2}-y ^{2}=y ^{2}\]

or

\[x ^{2}+y ^{2}=4\]

Now you can integrate. Limits on r will be 0 to 2 (circle of radius 2). Theta will be between 0 and pi/2 (since it's in the first octant). And limits on z will just be the two surfaces (converted to cylindrical coordinates, of course). Hope this helps.

Answer 2

I got the integral\[\large \int_{0}^{\frac\pi2}\int_{0}^{2}\int_{r^2\sin^2\theta}^{8-r^2(\cos^2\theta+1)}8r^3\cos\theta\sin\theta dzdrd\theta\]

Answer 3

would it be possible to integrate using rectangular coordinates?
using limits
x -> 0 to sqrt(4-y^2)
y -> 0 to sqrt(z)
z -> 0 to 4

or am i way off here