Problem: Let $S$ the solid limited by the surfaces
\begin{align} x= \sqrt{y^2 + z^2 }, \quad x=\sqrt{\frac{y^2 + z^2}{3}}, \quad x=\sqrt{a} \end{align}
the value of parameter $a$ for the following equality to be met
\begin{align} \int \int \int_S z dV = \pi \end{align}
is $\sqrt{2}$
My answer:
By the solid symetry, the centroid is at $z=0$ (thinking that the solid has homogeneus density $\rho=1$) so never $\int \int \int_S z dV = \pi $. What am I seeing wrong?.
Converting to polar
$x = x\\ y = r\cos \theta\\ z = r\sin \theta$
$\int_0^{2\pi}\int_0^{\sqrt a}\int_{x}^{x\sqrt 3} r^2\sin\theta\ dr\ dx\ d\theta$
I agree that will evaluate to 0.