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 $\pi/4$.

$$
I = \int\limits_D \frac{x^2}{x^2+y^2}\, dV
$$
$D$: $1 \leq x^2+y^2+z^2 \leq 2$ (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}
