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

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

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

So, I'm not really used to these kinds of problems, but here's what I got so far. Starting from the surface area function over cartesian coordinates, to change the variable of double integrals first we calculate the Jacobian as such:


$$ \frac{\partial f}{\partial r} = \cos\theta, \frac{\partial f}{\partial \theta} = -r\sin\theta, \frac{\partial g}{\partial r} = \sin \theta, \frac{\partial g}{\partial \theta} = -r \cos \theta $$

$$J = \left\lvert (\cos\theta)(-r \cos \theta) - (-r\sin\theta)(\sin \theta) \right\rvert = r$$

Not too much trouble there. But when I proceed to actually change the variables on the double integral, I don't know how to deal with partial derivatives of $f$. If I were to substitute using something like $x=r\cos \theta$ and $y=r\sin \theta$, I'd do something like this...

$$\iint_D \sqrt{ \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2 + 1} \,dA$$

$$\iint_Q \sqrt{ \left(\frac{\partial f}{\partial x}(r\cos \theta, r\sin \theta)\right)^2 + \left(\frac{\partial f}{\partial y}(r\cos \theta, r\sin \theta)\right)^2 + 1} \,r\,dr\,d\theta$$

Does that even make any sense? How would you proceed to do something like this? Thanks in advance.

  • $\begingroup$ Be careful: you’re using two different $f$’s and $g$’s above: $x$ is not equal to $f(r,\theta)$. It’ll be potentially less confusing to write $x=x(r,\theta)=r\cos\theta$ and $\frac{\partial x}{\partial r}=\cos\theta$ etc. You’re on the right track, though. You’ll need to use the chain rule to get farther. $\endgroup$
    – amd
    Commented Oct 15, 2015 at 10:24

2 Answers 2


Your setup amounts to a parametric representation $$(r,\theta)\mapsto{\bf x}(r,\theta)=\bigl(r\cos\theta,r\sin\theta, g(r,\theta)\bigr)$$ of the surface $S$ in question. Compute $${\bf x}_r=(\cos\theta,\sin\theta, g_r),\qquad{\bf x}_\theta=(-r\sin\theta,r\cos\theta,g_\theta)\ .$$ Then the surface area of $S$ is given by $$\omega(S)=\int_Q|{\bf x}_r\times{\bf x}_\theta|\>{\rm d}(r,\theta)= \int_Q\sqrt{r^2\bigl(1+g_r^2(r,\theta)\bigr)+g_\theta^2(r,\theta)}\>{\rm d}(r,\theta)\ .$$


Hint: Expand $\frac{\partial g}{\partial r}$ and $\frac{\partial g}{\partial\theta}$ using the chain rule:$$ \frac{\partial g}{\partial r} = \frac{\partial f}{\partial r}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial r}=f_x\cos\theta + f_y\sin\theta \\ \frac{\partial g}{\partial\theta} = \frac{\partial f}{\partial\theta}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial\theta}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial\theta}=f_y\;r\cos\theta-f_x\;r\sin\theta. $$Now, square these two expressions and look for a way to combine them that eliminates the coefficients of the partial derivatives of $f$.

  • $\begingroup$ Thanks a lot! This was enough to get me on the right track. $\endgroup$ Commented Oct 15, 2015 at 14:31
  • $\begingroup$ Quick question, how would you go about using that to prove the formula for the area of a surface of revolution resulting from rotating $y=f(x)$ around the $y$ axis? I mean, this one: $\int_0^a x \sqrt{1 + f'(x)^2} \,dx$ when $0 <= x <= a$. $\endgroup$ Commented Oct 15, 2015 at 14:48
  • $\begingroup$ For this special case, you have the coordinate transformation $(\bar x,\bar y)\mapsto(r,z)$ (I’ve used bars to avoid confusion with the Cartesian system in the original question). $g$ is a function of $r$ only, with $g(r)=f(\bar x)$ and $g_\theta=0$. Plug that into your formula or use a derivation parallel to that in Christian Blatter’s answer. $\endgroup$
    – amd
    Commented Oct 15, 2015 at 18:13
  • $\begingroup$ Oh, and you’re missing a factor of $2\pi$ in your surface of rotation formula. $\endgroup$
    – amd
    Commented Oct 15, 2015 at 18:29

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