Reduction to triple integral I have an integral 
$$I = \int_{V}(x^2 + y^2)\, dV, $$
where $V =  \{2ax \geqslant z^2 \} \cap \{ x^2 + y^2 \leqslant ax \}$.
Am I right, that 
$$I = \int_{0}^{a} \int_{-\sqrt{2ax}}^{\sqrt{2ax}} \int_{-\sqrt{a^2/4 - (x-a/2)^2}}^{\sqrt{a^2/4 - (x-a/2)^2}} (x^2 + y^2)\, dy\,dz\,dx?$$
It seems to me, that this is not really right, because I've tried to integrate it and my answer is far from correct, however I can't find where is the mistake.
 A: Your region $V$ is the intersection of the parabolic plane and the cylinder.
$\hspace{13em}$
So it is not bad to begin with cylindrical coordinates:
$$ (x, y, z) = (r \cos\theta, r\sin\theta, z) $$
Then $V$ is written as
$$ V = \{ 2ar \cos\theta \geq z^2 \} \cap \{ r \leq a \cos\theta \} $$
The above description provides a natural limits for $r$ and $z$. For $\theta$, we read out that $\cos\theta \geq 0$ and hence $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} $. So
\begin{align*}
\int_V (x^2 + y^2) \, dxdydz
&= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{a\cos\theta} \int_{-\sqrt{2ar\cos\theta}}^{\sqrt{2ar\cos\theta}} r^3 \, dz dr d\theta \\
&= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{a\cos\theta} r^{7/2} \sqrt{8a\cos\theta} \, dr d\theta \\
&= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{4\sqrt{2} a^5}{9} \cos^5 \theta \, d\theta \\
&= \frac{64\sqrt{2}a^5}{135}
\end{align*}
On the other hand, if you choose to stick to the Cartesian coordinates, you can still have an option:
\begin{align*}
\int_V (x^2 + y^2) \, dxdydz
&= \int_{0}^{a} \int_{-\sqrt{ax-x^2}}^{\sqrt{ax-x^2}} \int_{-\sqrt{2ax}}^{\sqrt{2ax}} (x^2 + y^2) \, dz dy dx \\
&= \int_{0}^{a} \int_{-\sqrt{ax-x^2}}^{\sqrt{ax-x^2}} (x^2 + y^2) \sqrt{8ax} \, dy dx \\
&= \int_{0}^{a} \frac{4\sqrt{2}}{3} x^2 (2x + a) \sqrt{a\smash[b]{(a-x)}} \, dx \\
&= \frac{64\sqrt{2}a^5}{135}.
\end{align*}
