Problem with the parametrisation of this surface integral I am facing troubles in understanding (read: "guessing") the correct way to parametrise this integral:
$$\int_{\Sigma} \dfrac{1}{\sqrt{1 + x^2 + y^2}}\ \text{d}\sigma$$
Where $\Sigma = \{(x, y, z)\in\mathbb{R}^3; x^2 + y^2 \leq 1; z = \sin^2(x^2+y^2)\}$
Is there a cool simple way to solve this?
I thought about the usual polar coordinated but it becomes rather strange.
Thank you!
Idea
Maybe it's going to be a real malarkey but I thought again about polar/cylindrical:
$$x = R\cos\theta$$
$$y = R\sin\theta$$
$$z = \sin^2(R^2)$$
Yet the ranges would become
$$R\in(-1, 1) ~~~~~~~ \theta \in (0, 2\pi) ~~~~~~~ z\in (0, 1)$$
Hence
$$\int_0^1 \text{d}z \int_{-1}^1 \frac{1}{\sqrt{1+R^2}}\ \text{d}R \int_0^{2\pi} \text{d}\theta = 4\pi\text{arcsinh}(1)$$
Yes/no?
 A: I think your miscalculating the integration limits of $z$. If you consider the change of coordinates $x = R\cos(\varphi)$, $y = R\sin(\varphi)$ and $z = z$, then the determinant of the corresponding Jacobian matrix is given by
\begin{align*}
|J| =
\begin{vmatrix}
\cos(\varphi) & -R\sin(\varphi) & 0\\
\sin(\varphi) & R\cos(\varphi) & 0\\
0 & 0 & 1
\end{vmatrix} = R\cos^{2}(\varphi) + R\sin^{2}(\varphi) = R
\end{align*}
Therefore we get that
\begin{align*}
\int_{\Sigma}\frac{1}{\sqrt{1+x^{2}+y^{2}}}\mathrm{d}\sigma = \int_{0}^{1}\int_{0}^{2\pi}\int_{0}^{\sin^{2}(R^{2})}\frac{R}{\sqrt{1+R^{2}}}\mathrm{d}z\mathrm{d}\varphi\mathrm{d}R
\end{align*}
Hopefully this helps.
EDIT
Let us consider the parametrization of $\Sigma$: $\displaystyle\sigma(u,v) = (u,v,\sin^{2}(u^{2}+v^{2})) = \left(u,v,\frac{1-\cos(2u^{2}+2v^{2})}{2}\right)$ whose domain is $D = \{(x,y)\in\textbf{R}^{2}:x^{2}+y^{2}\leq 1\}$. Then its partial derivatives are given by
\begin{align*}
\begin{cases}
\sigma_{u} = (1,0,2u\sin(2u^{2}+2v^{2}))\\\\
\sigma_{v} = (0,1,2v\sin(2u^{2}+2v^{2}))
\end{cases}
\end{align*}
Then we obtain the following integral:
\begin{align*}
\int_{\Sigma}\frac{1}{\sqrt{1+x^{2}+y^{2}}}\mathrm{d}\sigma & = \int_{D}\frac{1}{\sqrt{1+u^{2}+v^{2}}}\left\|\sigma_{u}\times\sigma_{v}\right\|\mathrm{d}u\mathrm{d}v\\\\
& = \int_{D}\frac{\sqrt{1 + 4(u^{2}+v^{2})\sin^{2}(2u^{2}+2v^{2})}}{\sqrt{1+u^{2}+v^{2}}}\mathrm{d}u\mathrm{d}v
\end{align*}
