# Evaluating a triple integral in spherical coordinates

I need to evaluate the integral $\int \int \int \frac{x^2}{x^2+y^2}$ over the region $D$ where $D = {(x,y)} : 1\leq x^2+y^2+z^2 \leq 2, z^2>=x^2+y^2$ and $z\leq 0$

So I tried converting to spherical coordinates, therefore $1\leq r\leq\sqrt{2}$ and $0\leq \theta \leq 2\pi$ and $0 \leq \phi \leq \pi/4$

as from the equation of the cone, $\cos{^2}{\phi} = \sin{^2}{\phi}$.

The integrand becomes $r^2 \sin{\phi} \cos{^2}{\theta}drd\theta d\phi$. So we now evaluate the integral and using wolfram alpha the answer was $\frac{\pi}{6}(5\sqrt{2}-6)$ however this is not the answer that my professor gave us to verify our result.

I am not sure where I have made a mistake. I assumed by the symmetry of the problem that the integral is the same for $\pm z$, hence $\phi$ goes from $0$ to $\pi/4$.

• Looks right to me. May 5, 2015 at 13:14
• @curo: I am curious if my result corresponds with the answer.
– mvw
May 5, 2015 at 13:15
• $\phi$ is usually the angle in the $x$-$y$ plane. $\theta$ is the one from the zenith downwards.
– mvw
May 5, 2015 at 13:16
• hmm it is not the same as the answer they gave, perhaps the lecturer was just wrong.
– curo
May 5, 2015 at 14:31

$$I = \int\limits_D \frac{x^2}{x^2+y^2}\, dV$$
$D$: $1 \leq x^2+y^2+z^2 \leq 2$ (spherical shell between $r=1$ and $r=\sqrt{2}$), $z^2>=x^2+y^2$ and $z\le 0$ (negative cone).
Going to spherical coordinates $dV = r \sin\theta\,dr\,d\theta\,d\phi$, $x = r \sin \theta \cos \phi$, $y = r \sin \theta \sin \phi$ gives