Calculating the integral of the Gaussian curvature I'd like to find $\int_{A} K dA$ for the surface $S = \{(x,y,z)\in\mathbb{R}^3 \vert x^2 + y^2 - z^{2/3} = 0, 1 < z < 8\}$
I've had an attempt but I'm fairly certain that this surface is diffeomorphic to a piece of the flat plane so the integral should be $2\pi$.
However, computing directly using the following parametrisation:
$x(r, \theta) = (r\cos(\theta), r\sin(\theta), r^3) $
Gives me: $E = 1+9r^4$, $F = 0$, $G = r^2$ for coefficients of the first fundamental form;
$L = \dfrac{6r^2}{(9r^6 + r^2)^{1/2}}$,   $M = 0$, $ N =  \dfrac{3r^4}{(9r^6 + r^2)^{1/2}}$, as coefficients of the 2nd fundamental form, so that:
$$K = \dfrac{18r^6}{(9r^6 + r^2)^2}$$
Then upon integrating:
$\int_A K dA = \int_0^{2\pi}\int_1^2 \dfrac{18r^6}{(9r^6 + r^2)^2} \sqrt{9r^6 + r^2 } dr d\theta $
I really have no idea where I've messed up here. It feels like the method is generally right, but I've just overlooked something. I would greatly appreciate any help spotting where I've messed up in my computation.
 A: First of all, I don't know why you expect an answer of $2\pi$. If $S$ and $S'$ are diffeomorphic, there's no reason their curvature integrals should match up. You'd need them to be isometric to arrive at that conclusion.
The correct formula for the Gaussian curvature is $K=18\dfrac{r^2}{(1+9r^4)^2}$ (which agrees with your formula after algebraic simplification). If you know the formula for Gaussian curvature in an orthogonal parametrization,
$$K=-\frac1{2\sqrt{EG}}\left(\Big(\frac{E_v}{\sqrt{EG}}\Big)_v+\Big(\frac{G_u}{\sqrt{EG}}\Big)_u\right),$$
it's quite easy to compute. But, better yet, you can easily evaluate the integral $\iint_S K\,dA$ in the case (as you have here) where $E$ and $G$ are functions of just one variable, $r$. Note that
\begin{align*}
\iint_S K\,dA &= \int_0^{2\pi}\int_1^2 -\frac1{2\sqrt{EG}}\Big(\frac{G_r}{\sqrt{EG}}\Big)_r\sqrt{EG}\,dr\,d\theta \\ &= -\frac12\int_0^{2\pi} \Big(\frac{G_r}{\sqrt{EG}}\Big)\Big]_{r=1}^{r=2}d\theta\\ &= \pi\Big(\frac{G_r}{\sqrt{EG}}\Big)\Big]_{r=2}^{r=1},
\end{align*}
since $G$ is independent of $\theta$.
