Compute the surface of $M=\{(x,y,z):2z=x^2+y^2,z<2\}$ 
Compute the surface of $M=\{(x,y,z):2z=x^2+y^2,z<2\}$.

EDIT
I've deleted my previous attempts, because they were wrong (as noted in the comments). While I have accepted an answer, I've written down my own approach here below. 
First we find a parametrization for our paraboloid. For $r\in(0,2),\theta\in(0,2\pi)$ we define
$$
\phi(r,\theta)=(r\cos\theta,r\sin\theta,r^2/2).
$$
The image of $\phi$ yields $M$, except for a set of measure zero, so that is ok. We want to compute the following integral:
$$
\int_MdV=\int_{(0,2)\times(0,2\pi)}V(D\phi).
$$
So we need the volume element $V(D\phi)=\sqrt{\operatorname{det}(D\phi^TD\phi)}$. We have
$$
D\phi(r,\theta)=\begin{pmatrix}
\cos\theta&-r\sin\theta\\
\sin\theta&r\cos\theta\\
r&0
\end{pmatrix},
$$
which yields
$$
D\phi^TD\phi=\begin{pmatrix}1+r^2&0\\0&r^2\end{pmatrix}.
$$
So we have $V(D\phi)=\sqrt{r^2(1+r^2)}=r\sqrt{1+r^2}$ (we can drop the absolute value for $r$, because $r>0$). At last, we compute the 2-dimensional volume:
$$
\int_{r=0}^2\int_{\theta=0}^{2\pi}r\sqrt{1+r^2}\,d\theta dr=2\pi\int_{r=0}^2r\sqrt{1+r^2}\,dr=\dfrac{2\pi}{3}(5\sqrt 5-1).$$
 A: In polar coordinates we have


*

*$0\le z \le 2$

*$0\le \theta \le 2\pi$

*$r(z)=\sqrt{2z}$


also note that


*

*$\frac{dr}{dz}=\frac1{\sqrt{2z}}\implies dr=\frac1{\sqrt{2z}}dz\implies ds=\sqrt{dz^2+dr^2}=\sqrt{\frac{1+2z}{2z}}dz$


then
$$S=\int_0^{2\pi}d\theta\int_0^2 r(z)\,ds=2\pi\int_0^2 \sqrt{2z}\sqrt{\frac{1+2z}{2z}}\,dz=2\pi\int_0^2 \sqrt{1+2z}\,dz$$
As an alternative by your parametrization the surface integral is
$$S=\iint_M|\vec \phi_r\times \vec \phi_\theta|\,dr\,d\theta$$
with


*

*$\vec \phi_r=(\cos \theta,\sin \theta,r)$

*$\vec \phi_\theta=(-r\sin \theta,r\cos \theta,0)$


and then


*

*$\vec \phi_r\times \vec \phi_\theta=(-r^2\cos \theta,-r^2\sin \theta,r)$

*$|\vec \phi_r\times \vec \phi_\theta|=\sqrt{r^4+r^2}=r\sqrt{1+r^2}$


and finally
$$S=\iint_Mr\sqrt{1+r^2}\,dr\,d\theta=\int_0^{2\pi} d\theta \int_0^2 r\sqrt{1+r^2}\,dr=2\pi\left[\frac13(1+r^2)^\frac32\right]_0^2=\frac{2\pi}3\left(5\sqrt 5 -1\right)$$
A: Noe That  the surface area of $$M=\{(x,y,z):2z=x^2+y^2,z<2\}$$ is the  surface of revolution  generated by rotating the region  in $xz$ plane bounded by $z=\frac {1}{2} x^2$, 
 $x=0$ , and $z=2$ about the $z-axis.$ 
Thus  the surface area is $$ S=\int _0^2 2\pi x ds = \int _0^2 2\pi x \sqrt {1+z'^2}dx=\\  \int _0^2 2\pi x \sqrt {1+x^2}dx =\\ \frac {2\pi}{3}(5\sqrt 5 -1)  $$
