Area of surface of revolution using polar coordinates

I've got the following problem for Calc III. If the surface area of a function $f(x,y)$ over a region $D$ is given in polar coordinates by the following double integral:

$$\iint_Q \sqrt{ r^2\left(1+\frac{\partial g}{\partial r}^2\right) + \left(\frac{\partial g}{\partial \theta}\right)^2} \,dr\,d\theta$$

$$g(r,\theta) = f(r \cos \theta,r \sin \theta)$$ $$Q = \{(r,\theta) \mid (r \cos \theta,r \sin \theta) \in D \}$$

How could I go about proving the formula for the area of a surface of revolution resulting from rotating the curve $y=f(x)$ about the $y$ axis using the integral above? $$A=\int_0^a x \sqrt{1 + f'(x)^2} \,dx$$

First of all you are wrong about second formula: $$y=f(x)\qquad a\le x\le b$$ $$A=2\pi \int_a^b x \sqrt{1 + f'(x)^2} \,dx$$ Imagine you have axises y and z instead x and y: $$z=f(y)$$ Now imagine you want area of the surface of revolution resulting from rotating the curve $z=f(y)$ about the $z$ axis. It means you should add third axis x to your system and observe that this surface has this equation: $$z=f(r)\qquad a\le r\le b\qquad 0\le \theta\le 2\pi$$ So by the first formula we have: $$\iint_Q \sqrt{ r^2({1 + f'(r)^2})+0} \,dr\,d\theta=2\pi \int_a^b r \sqrt{1 + f'(r)^2} \,dr$$