Evaluate $\int_0^1dx\int_0^{\sqrt{1-x^2}}dy\int_{\sqrt{x^2+y^2}}^{\sqrt{2-x^2-y^2}}z^2dz$ I need to evaluate this integral $$\int_0^1dx\int_0^{\sqrt{1-x^2}}dy\int_{\sqrt{x^2+y^2}}^{\sqrt{2-x^2-y^2}}z^2dz$$I know it's not hard, but somewhere I struggle. As I see it would be really helpful to use spherical coordinates. This is what I already did: $$\int_0^1dx\int_0^{\sqrt{1-x^2}}dy\int_{\sqrt{x^2+y^2}}^{\sqrt{2-x^2-y^2}}z^2dz=\int_0^{\frac{\pi}{2}}d\phi\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\theta\int_?^?r^4 \sin^2\theta 
 \cos^2\phi \space dr$$ where $x=r \cos\phi \cos\theta \space \space$, $y=r \sin\phi \cos\theta \space \space$ and $z=r \sin \theta$. 
Can anyone help me?
 A: 
Instead of going through the effort of working out the bounds for spherical coordinates, it may be less work to use cylindrical coordinates instead:
$$\begin{align}
\mathcal{I}
&=\int_{0}^{1}\mathrm{d}x\int_{0}^{\sqrt{1-x^{2}}}\mathrm{d}y\int_{\sqrt{x^{2}+y^{2}}}^{\sqrt{2-x^{2}-y^{2}}}\mathrm{d}z\,z^{2}\\
&=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\int_{0}^{1}\mathrm{d}r\,r\int_{\sqrt{r^{2}}}^{\sqrt{2-r^{2}}}\mathrm{d}z\,z^{2};~~~\small{\left[\left(x,y\right)=\left(r\cos{\theta},r\sin{\theta}\right)\right]}\\
&=\frac{\pi}{2}\int_{0}^{1}\mathrm{d}r\,r\int_{\sqrt{r^{2}}}^{\sqrt{2-r^{2}}}\mathrm{d}z\,z^{2}\\
&=\frac{\pi}{4}\int_{0}^{1}\mathrm{d}t\int_{\sqrt{t}}^{\sqrt{2-t}}\mathrm{d}z\,z^{2};~~~\small{\left[r^{2}=t\right]}\\
&=\frac{\pi}{4}\int_{0}^{1}\mathrm{d}t\int_{t}^{2-t}\mathrm{d}u\,\frac{u}{2\sqrt{u}};~~~\small{\left[z^{2}=u\right]}\\
&=\frac{\pi}{8}\int_{0}^{1}\mathrm{d}t\int_{t}^{2-t}\mathrm{d}u\,\sqrt{u}.\\
\end{align}$$
There rest of the calculation should be straightforward from this point.

