Write three integrals, one in Cartesian/rectangular, one in cylindrical, and one in spherical coordinates, that calculate the average of the function $f(x, y, z) = x^2 + y^2$ on the region $E$ in the first octant inside the sphere $x^2+y^2+z^2 = 9$, and above the cone $z=\sqrt{x^2+y^2}$.
The volume of $E$ is provided, $E = \frac{9\pi}{4}(2-\sqrt{2})$. Only the setup is needed, the integrals do not need to be evaluated.
I have the spherical and cylindrical integrals but I'm not quite sure of my bounds:
$\frac{1}{Vol(E)}\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{4}}\int_{0}^{3}\rho^4 \sin^3(\phi) d\rho d\phi d\theta$
$\frac{1}{Vol(E)}\int_{0}^{2\pi}\int_{0}^{\frac{3}{\sqrt{2}}}\int_{r}^{\sqrt{9-r^2}}r^3dz dr d\theta$
Best Answer
Yes your work is correct except the bounds of $\theta$. Please note that the region is in the first octant so $0 \leq \theta \leq \pi/2$.
In cartesian coordinates, note that at the intersection of the sphere and the cone,
$x^2 + y^2 = 9 - z^2 = 9 - x^2 - y^2 \implies x^2 + y^2 = 9/2$
$ \displaystyle \int_0^{3/\sqrt2} \int_0^{\sqrt{9/2-x^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{9-x^2-y^2}} (x^2 + y^2) ~ dz ~ dy ~ dx$