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? $$A=\int_0^a x \sqrt{1 + f'(x)^2} \,dx$$
Thanks in advance.