Calculate the volume between $2$ surfaces 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..
 A: A geometric view of the problem will be much of help to solve it.
One is a sphere of radius $\sqrt{8}$ centered at the origin.   
The other is a paraboloid of revolution, given by the revolution of $z=x^2/2$
around the $z$ axis, thus with the vertex at the origin.
The volume between the two is given by revolution around the $z$ axis
of the 2D area delimited by a parabola and a circle. 
I suppose you can compute that by "shells" or "washer" method.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}

Note that
  $\ds{0 < \rho/2 < \root{8 - \rho} \implies \rho < \color{red}{4}}$.

