0
$\begingroup$

Given I have some surface $ F(\rho, \theta)$ circular surface in the $xz$ plane, where $x = \rho sin(\theta),$ $z = \rho cos(\theta),$ $\rho : 0 \to \rho,$ $\theta: 0 \to 2\pi,$ $F = xi + zk.$

The way in which I would calculate the surface element $da$ in terms of the variables $\rho$ and $\theta$ is by computing

$|(\partial F /\partial \rho) ×(\partial F /\partial \theta)| d\rho d\theta$

Which gives me the standard result in polar/cylindrical coordinates that

$da = \rho d\theta d\rho$.

Intuitively, $\rho d\theta$ is the arc length of a sector of radius $\rho$ and angle $d\theta$ multiplied by some little $d\rho$, so you're finding the area of some tiny little strip.

I am now struggling to relate this to another derivation that I intuitively came up wit. Drawing the same sector diagram, the area of a sector would be $\frac{1}{2} \rho ^2 \theta$. The area of a sector of angle $d\theta$ would be $\frac{1}{2} \rho ^2 d\theta$. Now consider the sector of angle $d\theta$, but the radius is $(\rho - d\rho)$, overlaying them onto one another. Shouldn't the area of the sector of radius $\rho $ minus the sector of radius $(\rho -d\rho)$ give me the area of that same tiny little strip?

However, when doing the computation we get $$\begin{align} \frac{1}{2} \rho ^2 d\theta - \frac{1}{2} (\rho - d\rho)^2 d\theta, &= \rho d\theta d\rho - \frac{1}{2}(d\rho)^2 d\theta, \end{align}$$ which is wrong, apart from the very first term.

What have I done wrong? Clearly the correct way is to compute the cross product my surface to find $da$, yet my "intuition" gets the wrong answer.

$\endgroup$
1
  • $\begingroup$ was aware of that, but I chose to use x and z for my surface as I pictured the surface in the x z plain. it doesn't matter how I define the variables it is still the same as long as I'm consistent $\endgroup$ Nov 23, 2021 at 0:26

2 Answers 2

2
$\begingroup$

Your intuition is not given you the wrong answer, it is just based on a computation that is more suited for finite quantities, instead of infinitesimal. Notice that since $\textrm{d}\rho$ is very small (infinitesimal, in fact), the second term is negligible compared to the first one, and you can just throw it away.

This might seem weird as first, but it is due to the formal manipulation of infinitesimal quantities as if they were finite. As an example, let us try to derive in a similar manner the expression for the product rule. Consider two functions $x(t)$ and $y(t)$. We want to obtain the derivative of $xy$. Proceeding in a way similar to what you did, we find that $$\begin{align} \frac{\Delta (xy)}{\Delta t} &= \frac{(x + \Delta x) (y + \Delta y) - xy}{\Delta t}, \\ &= \frac{y \Delta x + y \Delta x + \Delta x\Delta y}{\Delta t}, \\ &= y \frac{\Delta x}{\Delta t} + y \frac{\Delta x}{\Delta t} + \frac{\Delta x\Delta y}{\Delta t}, \end{align}$$ which misses the correct result by a small amount. This is simply due to the fact that we computed it treating al quantities as finite, instead of bothering with the limits and stuff. Any Calculus textbook will show that the last term literally vanishes, while the other ones do not. Notice also that my example also has a geometrical picture: we can picture $x$ and $y$ as sides of a square and $\Delta x$ and $\Delta y$ as small increments to these sides, as done on Lecture 3 of these lecture notes.

In summary, your intuition is correct, but when dealing with infinitesimals formally we often get higher-order terms that should be neglected at the end. When doing things more rigorously (as in your first proof), they will vanish automatically, sometimes in such a subtle way we won't even notice they were ever there to begin with.

$\endgroup$
1
  • 2
    $\begingroup$ Good answer! I will add that it's easy to see which terms will vanish, just by looking at the order of the infinitesimals. A term like $d\rho d\theta$ in a double integral over $\rho$ and $\theta$ will potentially be finite, but a term like $d\rho^2 d\theta$ in a double integral can at most "get rid of" two infinitesimals by integration, and the third one will always remain in the result and in the limit take it to 0. $\endgroup$
    – noah
    Nov 23, 2021 at 0:46
-2
$\begingroup$

by definition, the surface element for a surface z=f(x,y) is given by

$dA=\left|\frac{\partial V}{\partial x}\wedge\frac{\partial V}{\partial y}\right|dxdy$

where V is the 3-vector defined as $V=(x,y,f(x,y))$ or in other words,

$dA=dx dy\sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}$

Now if one want to go to the cylindrical coordinates, you can do the coordinate change in the previous expression remembering that

$dxdy\to rd\theta dr$

or in other words

$dA=rdrd\theta\sqrt{1+\left(\frac{\partial f}{\partial r}\right)^2+\frac{1}{r^2}\left(\frac{\partial f}{\partial \theta}\right)^2}$

where the surface is now parametrized by $z=f(r,\theta)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.