Area under paraboloid $z=x^{2}+y^{2}$ with $z \geq 1$ Let surface be represented by $ \sigma (u, v) = (u, v,  u^{2}+v^{2}) $ with $u=r \cos \theta,v=r \sin \theta$ . Using differential geometry we get coefficients of first fundamental form as $E=1+4u^{2},F=4uv,G=1+4v^{2}$
Then the area will be $\int \int \sqrt{1+4(u^{2}+v^{2})}du dv$ 
Iam not clear with the next step in Presseley. He convert the variables and re write it as 
$\int \int \sqrt{1+4r^{2}} r dr d \theta$ How can we get this? 
 A: The surface is rotationally symmetric about the $z$ axis. Math often rewards us for using the symmetries present in our problem. Why not use polar (cylindrical) coordinates to begin with? The surface is given by $(r, \theta, r^2)$ for $0\leq r\leq 1, 0\leq \theta\leq 2\pi$.
Given a point $(r_0, \theta_0)$ and small positive numbers $\Delta r, \Delta \theta$, we have a region of points $(r, \theta)$ in the plane with $r_0\leq r\leq r_0+\Delta r$ and $\theta_0\leq \theta\leq \theta_0+\Delta\theta$. This region is roughly a rectangle with width $r_0\Delta\theta$ and length $\Delta r$.
The part of the parabola which is above this region is also nearly a rectangle. It also has width $r\Delta\theta$, but its length is longer, because it's not horizontal. By the Pythagorean theorem its length is $\sqrt{(\Delta r)^2 + (2r_0\Delta r)^2} = \sqrt{1 + 4r_0^2}\Delta r$.
So the area of that region of the surface is
$$
\sqrt{1 + 4r_0^2}\Delta r\cdot r_0\Delta\theta
$$
Summing up all these little parts, and letting the parts get smaller and smaller, the $\Delta$ turns into $d$, and we get the integral
$$
\iint\sqrt{1 + 4r^2}\,r\,dr\,d\theta
$$
One can, of course, get from your integral to this via a change in coordinates. You are, from some point on, expected to just know that $du\,dv$ in a cartesian coordinate system translates to $r\,dr\,d\theta$ in a polar coordinate system. But I prefer to not go through that hassle if I don't have to. It is one additional, unnecessary step.
