I'm trying to understand the proof for Green's Theorem and I've stumbled upon a few problems.

In my notes, it says that:

If $E$ is a simple (flat?) surface in $\mathbb{R}^2$ (I've been trying to find the right English terminology but I'm having trouble as it's not my first language, so I'm not sure if it's correct.), then $\partial E$ is its edge which is a piecewise smooth curve.

I don't really understand this conclusion.

Another thing I'm struggling with is the following:

If we have a scalar field Q (i.e. a function from $\mathbb{R}^2$ to $\mathbb{R}$) then $\frac{\partial Q}{\partial x}$ is a surface in 3-D space. (Again, not sure about the terminology, sorry.)

I've watched these videos: https://www.khanacademy.org/math/calculus/line_integrals_topic/greens_theorem/v/green-s-theorem-proof-part-1 hoping they would clarify the proof, which it did, but I don't understand it as intuitively as I want to.

If anyone could help me understand these concepts intuitively it would be much appreciated!

  • 2
    $\begingroup$ It rather depends on the context. It looks to me as though $\partial E$ is being used to mean the boundary of $E$ - and the fact that $\partial E$ is piecewise smooth will depend on what is said about the nature of $E$ or its boundary when it is first described or defined. The fact that the boundary is piecewise smooth is likely to be part of the definition, because you need some regularity in order to be able to integrate around the boundary. $\endgroup$ Mar 9, 2014 at 17:41
  • $\begingroup$ I think you may be tripping over a notational issue - $\partial$ is actually used for two distinct (though related) concepts here. $\frac{\partial Q}{\partial x}$ denotes the partial derivative of the (multivariate) function $Q$ with respect to $x$, but $\partial E$, by itself, denotes the boundary of the surface $E$; there's no derivative involved at all. Indeed, $E$ doesn't even need to be differentiable to have a boundary - imagine the surface $E = M\times [0, \infty)$, where $M$ is (the boundary of) the Mandelbrot set; then the boundary of $E$ is just $M$ itself. $\endgroup$ Mar 9, 2014 at 17:44

1 Answer 1


This isn't rigorous but the only reason I am posting this is "...could help me understand these concepts intuitively...". So here goes.

$\partial E$ would be a 'small change in E'. What elementary object is it, that if you keep adding side by side, will produce a surface? A line of course. It could be a curved one too, but a line nevertheless. And therefore the edge of the surface, since you are only incrementing(or decrementing) the surface at the edge.

The requirement of $\partial$ here is, that it could be differentiated with respect any of the two axes(assuming a planar surface) and you get a choice of selecting along which direction you want to increment. Hope that helps.

I cannot help with the second question, since I have never actually come across this before. I just answered with whatever physical intuition I had about differentiation, and its relation to change.


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