Calculate the volume between $x^2+y^2+z^2=8$ and $x^2+y^2-2z=0$. I don't know how to approach this but I still tried something:
I rewrote the second equation as: $x^2+y^2+(z-1)^2=z^2+1$ and then combined it with the first one and got $2(x^2+y^2)+(z-1)^2=9$ and then parametrized this with the regular spheric parametrization which is:
$$x=\frac {1}{\sqrt{2}}r\sin \theta \cos \phi$$ $$y=\frac 1{\sqrt{2}}\sin\theta\sin\phi$$ $$z=r\cos\theta + 1$$
And of course the volume formula:
$$V(\Omega)=\int\int\int_{\Omega} dxdydz$$
But that led me to a wrong answer.. what should I do?
Else, I tried parametrizing like this: $x=r\cos t$, $y=r\sin t$. then $r^2+z^2=8$ and $r^2-2z=0$ giving the only 'good' solutions $r=2, z=2$ then $r\in[0,2]$ and $z=[\frac {r^2}2,\sqrt{8-r^2}]$ positive root because it's in the plane $z=2$.
giving $\int_{0}^{2\pi}\int_0^2\int_{\frac {r^2}2}^{\sqrt{8-r^2}}rdzdrdt.$ But still i god the wrong answer..
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\begin{align} V & \stackrel{\mrm{def.}}{\equiv} \iiint_{\mathbb{R}^{\large 3}}\bracks{x^{2} + y^{2} + z^{2} < 8} \bracks{z > {x^{2} + y^{2} \over 2}}\dd^{3}\vec{r} \\[5mm] & \underline{\mbox{Use Cylindrical Coordinates:}} \\ V & = \int_{0}^{2\pi} \int_{-\infty}^{\infty}\int_{0}^{\infty}\bracks{\rho^{2} + z^{2} < 8} \bracks{z > {\rho^{2} \over 2}}\rho\,\dd\rho\,\dd z\,\dd\phi \\[5mm] & = \pi\int_{0}^{\infty}\int_{0}^{\infty}\bracks{\rho + z^{2} < 8} \bracks{z > {\rho \over 2}}\,\dd\rho\,\dd z \\[5mm] & = \pi\int_{0}^{\infty}\int_{0}^{\infty} \bracks{{1 \over 2}\,\rho < z < \root{8 - \rho}}\,\dd z\,\dd\rho \\[5mm] & = \pi\int_{0}^{\infty} \bracks{{1 \over 2}\,\rho < \root{8 - \rho}} \int_{\rho/2}^{\root{8 - \rho}}\,\dd z\,\dd\rho \\[5mm] & = \pi\int_{0}^{\color{red}{4}} \pars{\root{8 - \rho} - {1 \over 2}\,\rho}\,\dd\rho = \bbx{{4 \over 3}\pars{8\root{2} - 7}\,\pi} \approx 18.0692 \end{align}