I read, in a physics textbook that is not always as rigourous as desirable, that the $z$ component (and analogously for the other two components) of the curl of a vector field $\mathbf{E}$ is equal to the limit $$(\nabla\times\mathbf{E})_z=\lim_{\Delta S\to 0}\frac{1}{\Delta S}\oint_\gamma\mathbf{E}\cdot d\boldsymbol{\ell}$$where $\gamma$ is the positively oriented frontier of the rectangle contained in the $xy$ plane whose sides, having length $\Delta x$ and $\Delta y$, are respectively parallel to the $x$ and $yx$ axis and where $\Delta S$ is the rectangle surface.

From a recent answer I have got here from user251257, whom I thank again, I know that$$\lim_{ \substack{\sqrt{\sum_{i=1}^nh_i^2}\to0\\\prod h_i\ne0}} \frac{1}{∏h_i}\int _{x_{0,n}}^{x_{0,n}+h_n}...\int _{x_{0,1}}^{x_{0,1}+h_n}f(x_1,...,x_n)‌dx_1...dx_n=f(x_{0,1},...,x_{0,n})$$and, from that and Gauss-Green's theorem, I easily infer that $$(\nabla\times\mathbf{E})_z=\lim_{\sqrt{(\Delta x)^2+(\Delta y)^2}\to 0}\frac{1}{\Delta S}\oint_\gamma\mathbf{E}\cdot d\boldsymbol{\ell}$$but I am not able to prove my book's formula, if it really holds. Could anybody confirm whether $(\nabla\times\mathbf{E})_z=\lim_{\Delta S\to 0}\frac{1}{\Delta S}\oint_\gamma\mathbf{E}\cdot d\boldsymbol{\ell}$ holds and prove it if it does? I $\infty$-ly thank you for any answer!

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    $\begingroup$ Reading between the lines, is the question whether $\lim_{\Delta S\to 0}$ is the same as $\lim_{\sqrt{(\Delta x)^2+(\Delta y)^2}\to 0}$? $\endgroup$
    – mickep
    Sep 18, 2015 at 11:29

1 Answer 1


The question amounts to "is it OK to replace $\Delta S \to 0$ with $\sqrt{(\Delta x)^2 + \Delta y)^2} \to 0$," since everything else is the same. This is in the context of rectangle with sides $\Delta x$ and $\Delta y$, and $\Delta S$ representing (presumably) the area of the rectangle.

One problem here is that taking a limit as multiple things go to zero needs a very clear definition. Is it a limit as $\Delta x$ goes to zero, followed by a limit as $\Delta y$ goes to zero? The other order? Something else? Things like the limit with $\Delta S$ going to zero are even messier. Is it OK, in that case, to take a limit, as $t \to 0$, of a rectangle with sides $\Delta x = t^2 $ and $\Delta y = \frac{1}{t}$, so that the resulting rect has area $t$, which is going to zero?

In short: without knowing very clearly what the two limits really mean, there's no hope for you to prove they're equal.

The limit involving a square-root going to zero is pretty unambiguous, for that limit cannot go to zero without each of $\Delta x$ and $\Delta y$ going to zero. It says, roughly: for any $\epsilon > 0$, there's a number $\delta$ such that $\sqrt{(\Delta x)^2 + (\Delta y)^2}) < \delta$ implies that the thing in the limit expression differs from the desired limit by no more than $\epsilon$. If I were faced with such a thing to prove, I'd first observe that when the square root is less than delta, so are each of $|\Delta x|$ and $|\Delta y|$, and then do some algebraic fiddling with the stuff inside the limit using these two bounds.

But what does the limit with the $\Delta S$ actually mean? If it allows for the possibility that $\Delta x$ can be large as long as $\Delta y$ is much smaller, then I think that you're not going to be able to prove the thing you want to prove.

So while this doesn't answer the question directly --- "Can you show me how to prove this?" --- it does make clear what clarification is needed before someone like me might attempt that: I need to know a clear interpretation of what meaning you ascribe to the limit that involves $\Delta S$.

Here's a possibly useful example.

Consider the function $$ A(h, p) = p h^2 $$

The limit of $A(h, p)$ as $\sqrt{h^2 + p^2}$ goes to zero is in fact zero.

But if you instead ask "what's the limit of $A(h, p)$ as the product $hp$ goes to zero?", you have to find a limit $L$ with the property that for any sequence $(h_i, p_i)$ such that $\lim_{i \to \infty} h_i p_i = 0$, we have $$ \lim_i A(h_i, p_i) = L. $$

Let's look at two such sequences:

  1. $h_i = p_i = \frac{1}{i}$. In this case, the value of $A(h_i, p_i)$ is $\frac{1}{i} \left(\frac{1}{i}\right)^2$, which clearly goes to zero, so the limit (as $i \to \infty$) is zero.

  2. $p_i = \frac{1}{i^2}, h_i = i$. Then $p_i h_i = \frac{1}{i}$, which goes to zero. But $A(p_i, h_i) = \frac{1}{i^2} i^2 = 1$, so the limit in this case is 1.

Since these two limits on $i$ differ, there is no general limit as $p\cdot h \to 0$.

In short, as the area of the rectangle goes to zero, there is no limit, even though there is a limit as "both sides go to zero", so to speak.

  • $\begingroup$ You do answer me: the limit isn't defined for the surface approaching 0, while, if we intend that the diagonal approuaches 0, then the expression makes sense as I said in the original post. Thank you so much! $\endgroup$ Sep 18, 2015 at 11:37

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