Integration over the surface of a torus $$\boldsymbol{r}(\theta,\psi)=[(2+\cos(\theta))\cos(\psi), (2+\cos(\theta))\sin(\psi), \sin(\theta)]$$
For:
$$S = \lbrace\boldsymbol{r}(\theta,\psi)|\theta,\psi\in[0,2\pi]\rbrace$$
Evaluate:
$$\iint_S\frac{1}{\sqrt{x^2+y^2}}dA$$
 A: Up to Wiki the integral under consideration can be written as a double integral in such way: $$J:=\iint_S\frac{1}{\sqrt{x^2+y^2}}\,dA=\iint_S\frac{1}{\sqrt{((2+\cos(\theta))\cos(\psi))^2+((2+\cos(\theta))\sin(\psi))^2}}dA=$$ $$\iint_{\theta \in [0,2\pi],\\ \psi \in [0,2\pi]}\frac{1}{\sqrt{(2+\cos(\theta))^2}}\sqrt{\left(\frac {\partial(x,y)}{\partial(\theta,\psi)}\right)^2+\left(\frac {\partial(y,z)}{\partial(\theta,\psi)}\right)^2+\left(\frac {\partial(z,x)}{\partial(\theta,\psi)}\right)^2}\,d\psi\,d\theta. $$ Next, $$ \left(\frac {\partial(x,y)}{\partial(\theta,\psi)}\right)^2+\left(\frac {\partial(y,z)}{\partial(\theta,\psi)}\right)^2+\left(\frac {\partial(z,x)}{\partial(\theta,\psi)}\right)^2=$$ $$\left( -  \sin( \psi ) ^2\sin(
\theta ) \cos ( \theta ) -\sin ( \theta )
  \cos ( \psi ) ^2\cos ( \theta
 ) -2\,  \sin ( \psi )  ^2\sin
 ( \theta ) -2\,\sin ( \theta ) \cos
 ( \psi )  ^2 \right) ^2+
 $$
 $$(-(2+\cos(\theta))\cos(\psi)\cos(\theta))^2+(-\cos(\theta)(2+\cos(\theta))\sin(\psi))^2=$$ $$ \cos ( \theta )   ^2+4\,\cos \left( \theta
 \right) +4=(2+\cos(\theta))^2.
 $$ Thus, we have the happy end $$J= \int_0^{2\pi} \int_0^{2\pi}1\,d\psi\,d\theta=4\pi^2.$$
