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..
Best Answer
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.