# 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.

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@Arturo: It was supposed to be a "+". Thanks for editing it, I will take a look at the text and apply that in the future. Thanks! – Sir Winford Apr 27 '11 at 3:31

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.

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In fact you can immediately see using dimensional analysis that your expression $$\iint\limits_D \sqrt{1 + \left(\frac{\partial f}{\partial r}\right)^2 + \left(\frac{\partial f}{\partial\theta}\right)^2}r\,dr\,d\theta$$ has to be wrong. Remember that the surface has the unit of [length]$^2$, $r$ and $f$ are of unit [length], and $\theta$ is dimensionless.

This is a physicist way of saying that if you scale your Cartesian coordinate via $x'=\lambda x$, $y'=\lambda y$, $z'=\lambda z$ the the surface should scale like $\lambda^2$.

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