Calculating $\iiint_E\sqrt{3x^2+3z^2}\,dV$ 
Evaluate the following integral :$$\iiint_E\sqrt{3x^2+3z^2}\,dV$$where $E$ is the solid bounded by $y=2x^2+2z^2$ and the plane $y=8$.

I put $x=r\cos\theta$ and $z=r\sin\theta$ so, $y=2r^2\implies8\leq y\leq2r^2$. Finding limits of $r$ I got $0\leq r\leq2\csc\theta$ and for $\theta$ I got $\tan^{-1}4\leq\theta\leq\pi-\tan^{-1}4$.
$$\begin{align}\iiint_E\sqrt{3x^2+3z^2}\,dV &=\int_{\tan^{-1}4}^{\pi-\tan^{-1}4}\int_0^{2\csc\theta}\int_8^{2r^2}\sqrt{3}\cdot r\ dy \ r\ dr\ d\theta\\&=2\sqrt3\int_{\tan^{-1}4}^{\pi-\tan^{-1}4}\int_0^{2\csc\theta}r^2(r^2-4)dr\ d\theta\\&=2\sqrt3\ 8^3\int_{\tan^{-1}4}^{\pi-\tan^{-1}4}\csc^3\theta\bigg(\dfrac{8^2\csc^2\theta}{5}-\dfrac{4}{3}\bigg)d\theta\end{align}$$
Now, this is ugly to get through, please help, is there any efficient way to carry out this problem?

We do not have to convert to other coordinate system.  Since for a fixed value of y, the cross section is a circle with radius of $\sqrt{\frac{y}{2}}$.  Then the integral becomes
$$\int_{y=0}^{8} \int_{x=-\sqrt{\frac{y}{2}}}^{\sqrt{\frac{y}{2}}} \int_{z=-\sqrt{\frac{y}{2}-x^2}}^{\sqrt{\frac{y}{2}-x^2}} dz dx dy$$
$$= \int_{y=0}^{8} \frac{\pi}{2} \frac{\sqrt{3}}{\sqrt{2}} y^{\frac{3}{2}} dy$$
$$= \frac{128\sqrt{3}\pi}{5}$$
