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 $\pi/4$.

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

1 Answer 1


$$ I = \int\limits_D \frac{x^2}{x^2+y^2}\, dV $$

$D$: $1 \leq x^2+y^2+z^2 \leq 2$ (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

\begin{align} I &= \int\limits_D \frac{r^2 \sin^2 \theta \cos^2 \phi}{r^2 \sin^2 \theta} \, dV \\ &= \int\limits_D \cos^2 \phi \, r \sin\theta\,dr\,d\theta\,d\phi \\ &= \int\limits_0^{2\pi} \cos^2 \phi \, d\phi \int\limits_1^\sqrt{2} r\, dr \int\limits_{3\pi/4}^\pi \sin\theta\, d\theta \\ &= \frac{1}{2}\left[\phi + \cos \phi \sin \phi \right]_0^{2\pi} \frac{1}{2}\left[ r^2 \right]_1^\sqrt{2} \left[ -\cos \theta\right]_{3\pi/4}^{\pi} \\ &= (\pi)(1/2)(1 -1/\sqrt{2}) \\ &= \frac{\pi}{2} \left(1 - \frac{1}{\sqrt{2}} \right) \end{align}


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