What is the correct result to this surface integral?
$$\int_{\Sigma}x^2\,d\sigma$$ $\Sigma=\lbrace(x,y,z)\in \mathbb{R}^3 : (x,y)\in A,z=\operatorname{arctg}(y/x) \rbrace$ , with $A=\lbrace (x,y)\in \mathbb{R^2}:x\geq 0 , y\geq 0, 1\leq x^2+y^2\leq 2\rbrace$.
Thanks in advance.
Recall that $$\int_\Sigma f\,\mathrm d\sigma = \iint_A f(\sigma(x, y))\|N(x, y)\|\mathrm dx\mathrm dy,$$ with $\sigma \colon A \to \mathbb R^3$, $\Sigma = \sigma(A)$ and $N$ the normal vector.
In this case, since the surface is in the form $z = \sigma(x, y)$ the normal vector is simply $$N(x, y) = (-\partial_x f, -\partial_y f, 1) = \left(\frac{y}{x^2 + y^2}, -\frac{x}{x^2 + y^2}, 1\right).$$ Hence $$\|N(x, y)\| = \sqrt{\frac{1 + x^2 + y^2}{x^2 + y^2}}.$$
The integral becomes $$\int_A x^2\sqrt{\frac{1 + x^2 + y^2}{x^2 + y^2}}\, \mathrm dx\mathrm dy$$ which in polar coordinates is: $$\int_0^{\pi/2}\int_1^\sqrt 2 \rho^2\cos^2\theta\sqrt{1 + \rho^2}\,\mathrm d\rho\mathrm d\theta = \frac\pi{32}\Big(5\sqrt6 - 3\sqrt2 + \operatorname{arcsinh}1 - \operatorname{arcsinh}\sqrt2\Big)$$