Evaluate $\iiint_V(x+zy)^2dxdydz$ where V is bounded by $x^2+ y^2 = 2z, z = 2 $

Question

Evaluate $\iiint_V(x+zy)^2dxdydz$ where V is bounded by $x^2+ y^2 = 2z, z = 2 $.

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$x^2+ y^2=4$ for $z =2$ so $z\in{[0,2]}$ and $y\in{[0,{\sqrt{4-x^2}}]}$. Is it correct and how can I find $x$ interval?

Answer 1

It's better to use cylindrical coordinates here. Your equations become $r^2=2z$ and $z=2$. So, $z$ lies in the interval $[0,2]$ and, for each $z$, $r\leqslant\sqrt{2z}$. Therefore, compute the integral$$\int_0^{2\pi}\int_0^2\int_0^{\sqrt{2z}}r\bigl(r\cos(\theta)+zr\sin(\theta)\bigr)^2\,\mathrm dr\,\mathrm dz\,\mathrm d\theta$$