Solving triple integral I have to solve the integral $$ \int_{0}^{R}dz \int_{0}^{R} dy \int_{0}^{\sqrt{R^2-y^2}}dx$$
 of the function $x^2+y^2+z^2$ .
My problem is that $dz$ and $dy$ are not at the end, so I cannot use other coordinate transformations, e.g. in cylindric $dxdydz = \rho d\rho d\phi dz$
How do I deal with that?
 A: When calculating
$$\int_{0}^{R}dz \int_{0}^{R} dy \int_{0}^{\sqrt{R^2-y^2}}dx(x^2+y^2+z^2),$$
you can integrate consecutively with respect to each variable, no need to integrate with respect to $dx\,dy\,dz$. In your example you obtain
$$\int_{0}^{R}dz \int_{0}^{R} dy \int_{0}^{\sqrt{R^2-y^2}}dx(x^2+y^2+z^2) =\int_{0}^{R}dz \int_{0}^{R} dy \left(\frac{x^3}{3}\Big|_{x=0}^{x=\sqrt{R^2-y^2}}+(y^2+z^2) \sqrt{R^2-y^2} \right)$$ and then continue with integrating with respect to $dy$ and $dz$.
Another approach would be to examine the region where you are integrating: $0\le z\le R$, $0\le x$, $0\le y$, $x^2+y^2\le R$, which strongly suggests the cylinder coordinates: $$x= r\cos \phi\\y=r\sin \phi\\z=z.$$
In these coordinates your integral becomes
$$\int_0^Rdz \int_0^R rdr\int_0^{\pi/2} (z^2+r^2)d\phi,$$which should be even easier to find.
A: Are you sure you are not losing yourself with not much? Note that $$ \int_{0}^{R}dz \int_{0}^{R} dy \int_{0}^{\sqrt{R^2-y^2}}dx\,(x^2+y^2+z^2)=\int_{0}^{R}\left(\int_{0}^{R} \left(\int_{0}^{\sqrt{R^2-y^2}}(x^2+y^2+z^2)dx\right) dy\right) dz,$$ and that $$\int_{0}^{R}\int_{0}^{\sqrt{R^2-y^2}}(x^2+y^2+z^2)dx\,dy=\iint_D(r^2+z^2)\,r\,dr\,d\theta,$$ where $D$ is the quarter of a disk defined in polar coordinates by $$D=\{(r,\theta)\mid 0\leqslant r\leqslant R,\,0\leqslant\theta\leqslant\tfrac\pi2\}.$$
