I have to evaluate the integral $$\iiint_{W} (x^2+y^2+z^2)^{3/2}dV$$ where $W$ is the region which lies in the first octant inside the sphere $$x^2+y^2+z^2=16$$ and bounded by the cones $$z=\pm\sqrt{x^{2}+y^{2}}$$
In spherical coordiantes, clearly $$0\leq\rho\leq4$$ and since it's in the first octant, then $$0\leq\theta\leq\frac{\pi}{2},\qquad\frac{\pi}{4}\leq\phi\leq\frac{\pi}{2}$$
Therefore, the integral will be $$\int_{0}^{\pi/2}\int_{\pi/4}^{\pi/2}\int_{0}^{4}\rho^{5}\sin(\phi)d\rho d\phi d\theta$$ which value is $$\frac{512\sqrt{2}\pi}{3}$$ Is it correct?