Calculate the integral of $f(x,y,z)=12xyz$ on $D=\{ (x,y,z) \in R^3 \mid 0<x,y, \text{ and } \sqrt{2}(x^2+y^2)<z<\sqrt{3-x^2-y^2})\}$

Question

I'm learning to solve this kind of problems at the Calculus class, but I couldn't figure out this one.

I think i might have to use cylindrical coordinate, but i couldn't parameterize it.

(Sorry for my bad English.)

Answer 1

If you know to use cylindrical coordinates, there is no further difficulty really.

$x = r\cos\theta, y = r\sin\theta, z = z\implies 0<\theta<\frac \pi 2,\,\sqrt{2}r^2<z<\sqrt{3-r^2}.$ This means you need to find the bounds for $r$ by solving the previous inequality:

$2r^4<3-r^2$ or $2r^4+r^2<3\iff (r^2-1)(2r^2+3)<0\implies 0<r<1$, since $r$ is positive by default. Now your integral is: $$I=\int_0^{\frac{\pi}{2}}\int_{0}^1\int_{\sqrt{2}r^2}^{\sqrt{3-r^2}}12r^3z\cos\theta\sin\theta dzdrd\theta.$$

Do not forget the extra $r$ from the Jacobian when changing to cylindrical coordinates.