Find the volume of $$z^2\;\leq\;y\;\leq\;2-2x^2-z^2,\qquad x\geq0.$$
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!
We have
$$\iiint\limits_{\text{Solid}}{\text dx\text dy\text dz}=\int_{-1}^1\, dz\int_0^{\sqrt{1-z^2}} \, dx\int_{z^2}^{2-2x^2-z^2}\, dy=\frac{\pi}2$$
and by
$x=\rho \sin \theta$
$z=\rho \cos \theta$
$y=y$
$$\int_0^\pi\text d\theta\displaystyle\int_0^1\rho\text d\rho\displaystyle\int_{\rho^2\cos^2\theta}^{2-\rho^2-\rho^2\sin^2\theta}\text dy=\frac{\pi}2$$
The result is the same but in your set up $x$ and $z$ have been exchanged.