The answer should be obvious to me but I am not quite certain I am getting the right idea.
It is asked to calculate the volume of $M=\{(x,y,z) : 2z=x^2+y^2, z<2\}$. So this gives me the following:
$0<z=\frac{x^2+y^2}{2}<2 \implies 0<x^2+y^2<4$. So let's put $r=\sqrt{x^2+y^2}$. Then $0<r<2$. And the integral in cylindrical coordinates then becomes:
$\int_M=\int_{z=0}^{z=2}\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=2}rdr d\theta dz = 8\pi$.
Is this the correct way?
Thanks all.
On
If you want the volume of the domain enclosed by the three dimensional surface $\;M\;$ between $\;0\le z\le 2\;$ , then the integral is
$$\int_0^{2\pi}\int_0^2\int_{r^2/2}^2r\,dz\,dr\,d\theta=2\pi\int_0^2r\left(2-\frac12r^2\right)\,dr=2\pi\int_0^2\left(2r-\frac12r^3\right)\,dr=$$
$$=\left.2\pi\left(r^2-\frac18r^4\right)\right|_0^2=2\pi\left(4-2\right)=4\pi$$
What you calculate with
$$\int_M\,\mathrm dV=\int_{z=0}^{z=2}\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=2}rdr d\theta dz $$
Is the volume of the domain between the $\;xy\,-$ plane and the surface $\;M\;$ itself (i.e, between the floor and the bowl, not the volume enclosed by the bowl itself!)
As you wrote it it is a little confusing, but I think they meant the first one...
Check the bound for $r$ we should have $r$ from $0$ to $f(z)$, that is
$$\int_0^{2\pi}\,d\theta\int_0^2\,dz\int_{0}^{\sqrt{2z}}r\,dr=2\pi\int_0^2z\,dz=4\pi$$