<s>[INST] Find $\iiint (x^2+y^2+z^2)~dV$, above the cone $z=\sqrt{3(x^2+y^2)}$, inside the sphere $x^2+y^2+z^2\leq a^2$. 
Find
  $$\iiint (x^2+y^2+z^2)~dV$$
  above the cone $z=\sqrt{3(x^2+y^2)}$, and inside the sphere $x^2+y^2+z^2\leq a^2$.

I think that the solution for this problem is following:
Use cylindrical coordinates:
$x = r\cdot \cos(\theta)$
$y = r\cdot \sin(\theta)$
$z = z$
and we will get:
$$\int_0^a \int_0^{\pi} \int_0^{a/2} (r^2+z^2)r dr d\theta dz$$
Is this approach correct?
 [/INST] Hint: Alternative approach is using the spherical coordinates for this problem:
$$\int_0^a \int_0^{2\pi} \int_{0}^{\frac{\pi}{3}} r^4\sin\varphi d\varphi d\theta dr=\color{blue}{\dfrac{\pi}{5}a^5}$$
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