The statement of Green’s theorem is not particularly illustrative of the ideas driving it. We would try to gain some intuition by considering a simple example: finding the area bounded by a curve in the plane.

We would assume some knowledge of line integrals. Consider the following special case of Green’s theorem.

Theorem: Given some positively oriented, piecewise smooth, simple closed curve in a plane. Let D be the region bounded by C. Then the area of D is

$\text{Area of }D = \iint_D dx dy = - \int_C y dx$

This looks awfully similar to finding the area under curves. Recall that

Lemma: Given a function $$y(x)$$, the area under the curve $$\{(x, y(x)) \, | \, x \in [a,b]\}$$ is $$\int_a^b y(x) dx$$

Can we use that idea alone to prove the theorem? Yes! Given some nice looking curve $$C$$, we can split it into two parts. Express each part as some plot of some functions $$f(x), g(x)$$.

We could see that the area is now $$\int_a^b (f(x) - g(x)) dx$$ which is the same as $$- \int_C y dx$$!

We could’ve done this by considering $$\int_C x dy$$ instead. We would’ve had to taken a bit more care. In particular it would be nicer to split the curve into $$C_1$$ and $$C_2$$ so we get nice single-valued $$f_i(y)$$ and $$g_i(y)$$. (The reason we can split the curve is that the sum of the areas bounded by $$C_1$$ and $$C_2$$ is the area bounded by $$C_1$$) The driving principles are essentially the same.

Fundamentally, counting the area of a domain $$D$$ should be a double integral $$\iint_D dx dy$$. It’s only when we make some choice (and then using Green’s theorem or otherwise) would we have been able to reduce it into a single integral.

## Generalisation

For the actual statement of Green’s theorem, one could refer to a lecture by MIT OpenCourseWare or a sketch of proof associated with the lecture (pdf).

Perhaps I would add an informal comment that a slight reformalisation of Green’s theorem would be

$\int_C \bigg( \int M(x,y) dy \bigg) dx = \iint_D -M(x,y) dx dy$

which highlights the fact that in both sides of the equation two integrations had to be done: the number of integrations we do is the same. One may notice its similarities with Fubini’s theorem.

Here, we’ve made a choice of direction of integrating along the $$y$$-axis first, then the $$x$$-axis. We could’ve for example stated the following instead, which integrates along the $$x$$-axis first.

$\int_C \bigg( \int N(x,y) dx \bigg) dy = \iint_D N(x,y) dx dy$

By rotating the plane the direction could be arbtiary. We could’ve even choosen to use other coordinates (such as polar coordinates).

The advantage of making a choice is that we could more easily relate the area integral with integrals over $$\mathbb{R}$$. The disadvantage is that the formulation is less elegant / generalised than the Generalized Stokes theorem.