Computing a surface area using cylindrical coordinates $z = f(r,\theta)$, where $(r,\theta)$ varies through a region $D$ on the $r\theta$-plane, $r$ non-negative. 
I need to show that the surface area of the surface is given by 
$$\underset{D}{\iint} \sqrt{1 + \left(\frac{\partial f}{\partial r}\right)^2 + \frac{1}{r^2}\left(\frac{\partial f}{\partial\theta}\right)^2}r\,\mathrm dr\,\mathrm d\theta.$$
I can solve this problem except for the part including $(1 / r^2 )$.  I have no idea where that component comes from.  Any help is appreciated.
 A: I'm not sure how you got only part of the formula... perhaps you made a calculation error?  Did you remember that $x^2 + y^2 = r^2$ and not $x^2 + y^2 = r$?
At any rate, I assume that you're starting with the formula
$$S = \iint_D \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left( \frac{\partial f}{\partial y}\right)^2}\,dx\,dy$$
and going from there.
Naturally, you'll need to use the chain rule for partial derivatives
$$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x}$$
$$\frac{\partial f}{\partial y} = \frac{\partial f}{\partial r}\frac{\partial r}{\partial y} + \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial y},$$
while in the process getting rid of any $x$- and $y$-terms via the formulas
$$x = r\cos \theta$$
$$y = r\sin \theta.$$
Finally, the change of variables formula lets you write (formally) $$dx\,dy = r\,dr\,d\theta,$$
so that should account for the final $r$-term outside of the square root.
I'll leave the details of the calculation up to you.
