**Find the volume of** $$z^2\;\leq\;y\;\leq\;2-2x^2-z^2,\qquad x\geq0.$$

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I find the intersection of the curves: $$\begin{matrix}z^2&=&2-2x^2-z^2&&\Rightarrow&&2x^2+2z^2&=&2&&\Rightarrow&&x^2+z^2&=&1,\end{matrix}$$ and as $x\geq0$ then we have a semicircle of radius $1$ in the $xz$-plane. So the volume becomes:

$$\begin{matrix}V&=&\displaystyle\iiint\limits\_{\text{Solid}}{\text dx\text dy\text dz}&=&\displaystyle\iint\limits\_{P\_{xz}}{\text dx\text dz}\displaystyle\int\limits\_{z^2}^{2-2x^2-z^2}{\text dy}.\end{matrix}$$ Using cylindrical coordinates we have that the last integral becomes

$$\begin{matrix}
&&\displaystyle\int\_0^\pi\text d\theta\displaystyle\int\_0^1\rho\text d\rho\displaystyle\int\_{\rho^2\sin^2\theta}^{2-\rho^2-\rho^2\cos^2\theta}\text dz&(1)\\\\
&=&\displaystyle\int\_0^\pi\text d\theta\displaystyle\int\_0^1\rho\big[2-\rho^2\underbrace{-\rho^2\cos^2\theta-\rho^2\sin^2\theta}\_{-\rho^2}\big]\text d\rho&(2)\\\\
&=&\displaystyle\int\_0^\pi\text d\theta\displaystyle\int\_0^1[2\rho-2\rho^3]\text d\rho&(3)\\\\
&=&\displaystyle\int\_0^\pi\text d\theta\left.\left[\rho^2-\dfrac{\rho^4}2\right]\right|\_0^1&(4)\\\\
&=&\dfrac 12\displaystyle\int\_0^\pi\text d\theta&(5)\\\\
V&=&\boxed{\dfrac\pi2}.&(6)
\end{matrix}$$

Is this correct?

Thank you!