<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] $\mathbf{HINT}$: First we have  $\sqrt{3(x^2+y^2)}\leq z \leq \sqrt{a^2-x^2-y^2} $.
Second solve system: $z=\sqrt{3(x^2+y^2)}$, $x^2 + y^2 + z^2 = a^2 $. Solution: $D:x^2+y^2 = \frac{a^2}{4}$.
So, $$\iiint (x^2+y^2+z^2) dV = \iint_{D} \int_{\sqrt{3(x^2+y^2)}}^{\sqrt{a^2-x^2-y^2}} (x^2+y^2+z^2) dz dxdy,$$
then we use cylindrical coordinates: $x=r\cos(\phi), y=r\sin(\phi),z=z$ and we have
$$\int_{0}^{2 \pi} \int_{0}^{a/2} \int_{\sqrt{3r^2}}^{\sqrt{a^2-r^2}} (r^2+z^2)r dz dr d\phi=...=\frac{\pi a^5}{5}$$
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