I have a problem of which the textbook offers no clarification. The problem states
Evaluate the surface integral of f(x,y) over the rectangle $0 \le x \le a, 0 \le y \le b$ for the function
$$f(x,y) = \frac{x}{x^2+y^2}$$
Instinctively, I want to evaluate the integral by
$$S=\int_{0}^{b}\int_{0}^{a}\frac{x}{x^2+y^2}dxdy$$
However, when searching about surface integrals I found that they are evaluated by
$$S=\int\int_{D}{}f(x,y,g(x,y))\sqrt{(\frac{\partial g}{\partial x})^2+(\frac{\partial g}{\partial y})^2+1} dA$$
I have never seen this formula in class so I'm not too sure on whether this is what I should be using. Also, if I were to use the final integral where in my case $g(x,y)=0$ wouldn't my question reduce to my first integral?
Surface area of the surface $z =f(x,y) = \frac{x}{x^2+y^2}$ is given by
$$ S = \iint_{D} \sqrt{f_x^2+f_y^2+1} \,dA\\ f_x = \frac{y^2-x^2}{(x^2+y^2)^2}\\ f_y = \frac{-2xy}{(x^2+y^2)^2}\\ \sqrt{f_x^2+f_y^2+1} = \sqrt{2}\frac{1}{x^2+y^2}\\ $$
Converting to cylindrical coordinates,
$$ S =\sqrt{2}\left( \int_{0}^{\frac{\pi}{4}} \int_{a}^{\sqrt{a^2+b^2}} \frac{1}{r^2}r dr d\theta + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_{\sqrt{a^2+b^2}}^{b} \frac{1}{r^2}r dr d\theta\right)$$
$$S = \frac{\pi}{2\sqrt{2}} \log \left(\frac{b}{a}\right)$$