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I am trying to compare the line integral stated in Green's Theorem with the definition of a line integral. According to Wikipedia: $$ \oint_C(L dx+Mdy)=\int^b_af(\textbf{r}(t))|\textbf{r}'(t)|dt. $$ My intuition tells me that $\textbf{r}(t)=(x(t),y(t))$. Let $ds=\sqrt{dx^2+dy^2}$. Thus, $|\textbf{r}'(t)|dt=ds$, right? So if $C$ is a simple closed curve, is the expression below equivalent to the expression of the line integral in Green's Theorem? $$ \oint_C(L+M)ds $$ My question really is: what are $f$ and $\textbf{r}$ in the statement of Green's theorem? I was only able to recreate the statement by definition by letting $L$ and $M$ be independent of $y$ and $x$ respectively, as seen below: $$ \oint_CL(x(t))|\frac{dx}{dt}|dt+M(y(t))|\frac{dy}{dt}|dt. $$ The problem that I run into here is that $\partial L/\partial y$ and $\partial M/\partial x$ are then both 0.

Tom O'Shaughnessy
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  • Usually the statement of the Green Theorem equates the integral of some scalar $f=f\left(\mathbf{r}\right)$ over domain (area) $D$, with integral of $L dx + M dy$ along the boundary of that domain $\partial D$, which can be parameterized by $x=x(t),,y=y(t)$. You also need $\partial_x M - \partial_y L = f$. So I would expect to see a 2D integral on one side of equality, and a 1D on another. You have 2 one-dimensional integrals. Am I missing something? – Cryo Feb 23 '19 at 01:46
  • I did not include that side of the equality in the post because I did not believe it was relevant. I am only asking about the line integral side. – Tom O'Shaughnessy Feb 23 '19 at 01:49

2 Answers2

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You're right about what $\mathbf{r}$ is, but not about the structure of the integral itself.

The function we're actually integrating in the line integral side of Green's theorem is a dot product: $$\int_a^b \mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(t)\,dt$$ where $\mathbf{F}(x,y)=(L(x,y),M(x,y))$. By standard theory, this doesn't depend on exactly how we parametrize the curve.

That's the one-dimensional integral. You didn't say anything about the two-dimensional integral on the other side of the integral, so I'll assume you understand that part.

jmerry
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  • Can $f(\textbf{r}(t))$, in the definition of the OP, be written as $|\textbf{F}(\textbf{r}(t))|\cos(\theta(\textbf{r}(t)))$, where $\theta(\textbf{r}(t))$ is the angle between $\textbf{F}(\textbf{r}(t))$ and $\textbf{r}'(t)$? – Tom O'Shaughnessy Feb 23 '19 at 02:05
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    There is no such $f$. The point is to have a single function based on $L$ and $M$ which we will then evaluate at the points $r(t)$ on the curve. No scalar function can do that - because that angle is a function of the curve as a whole, not a single point. By taking a different curve, we can go through that point at a different angle, and we would then need a different number there. The vector formulation can handle that. A scalar function can't. – jmerry Feb 23 '19 at 02:23
  • Isn't $\theta(\textbf{r}(t_0))$ the angle at one point, $(x(t_0),y(t_0))$? A different curve would perhaps cause $\theta(\textbf{r}(t_0))$ to yield a different angle, would it not? – Tom O'Shaughnessy Feb 23 '19 at 02:46
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    Yes, and that means you'd need a different number for $f$ - at the same point. A function evaluated at a point has one value. That's what I was saying. – jmerry Feb 23 '19 at 02:49
  • Ah! Yes, I get it now. – Tom O'Shaughnessy Feb 23 '19 at 02:52
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You seem to be trying to represent a 1-form $\omega=L\left(\mathbf{r}\right) dx + M\left(\mathbf{r}\right) dy$ with a single scalar function $f=f\left(\mathbf{r}\right)$. This will probably not work for the same reason that you cannot represent a 2D vector (function) with a scalar function - you simply have more degrees of freedom in the former.

Cryo
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