Using spherical coordinates evaluate:
$ \displaystyle \int_0^3 \ \int_0^{\sqrt{9-y^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{18-x^2-y^2}} (x^2+y^2+z^2) \ dz \ dx \ dy $
From the rough sketch that i drew:
I find out that $\theta$ lies between $0$ and $\pi$/2
Then from the range of the integral of
$z$ = $\sqrt{x^2+y^2}$
$\phi$ = $\pi$/4
Thus, $\phi$ lies between $0$ and $\pi$/4
Lastly, for $\rho$
$z$ = ${\sqrt{18-x^2-y^2}}$ , $\rho$ is ${\sqrt{18}}$
Thus, $\rho$ lies between $0$ and ${\sqrt{18}}$
But my answer is wrong, can you please guide me since I think that my range of $\rho$ is wrong?
Bounds you have come up with are correct. You may have made a mistake somewhere in computation.
$I = \displaystyle \int_0^{\pi/2} \int_0^{\pi/4}\int_0^{\sqrt{18}} \rho^4 \ \sin \phi \ d\rho \ d\phi \ d\theta$
$ = \displaystyle \int_0^{\pi/2} \int_0^{\pi/4} \frac{972\sqrt2}{5} \sin \phi \ d\phi \ d\theta$
$ = \displaystyle \int_0^{\pi/2} \frac{972\sqrt2}{5} \left(1 - \frac{1}{\sqrt2}\right) \ d\theta$
$ \displaystyle = \frac{486 \pi}{5} (\sqrt2 - 1)$