Confusion with 2d Divergence theorem

Question

The 2d Divergence theorem says that $$\int_{\partial \Omega} P dy - Qdx\,\, {\color{red} =} \int_{\partial \Omega} (P,Q) \cdot n d \ell = \int_{\Omega} \nabla \cdot (P,Q) dA. $$ I am unsure why would the first equality be true. Let $ \Omega$ be a rectangle where the boundary is labelled as $R_1, R_2, R_3, R_4$ and oriented anticlockwise. Then \begin{align*} \int_{\partial \Omega} P dy - Qdx &= \int_{R_1} -Q dx + \int_{R_2} Pdy + \int_{R_3} -Q dx + \int_{R_4} Pdy. \end{align*} \begin{align*} \int_{\partial \Omega} (P,Q) \cdot n d \ell &= \int_{R_1} -Q dx + \int_{R_2} Pdy + \int_{R_3} Q dx + \int_{R_4} -P dy, \end{align*} since the normal is in the negative direction in the first and fourth integral. Since they differ on the last two term, I am not sure how can they be equal?

Answer 1

To sort out your specific example, taking into account the normal correctly, starting from $\int (P,Q) \cdot n d \ell$ you get $\int_{R_1} Q d \ell + \int_{R_2} -P d\ell + \int_{R_3} -Q d \ell + \int_{R_4} P d \ell$. This is because the normals are $e_2,-e_1,-e_2$ and $e_1$ respectively

The key now is that $dx$ and $dy$ are signed while $d \ell$ is not. So working from the other side, you have in each integral $dx=-d\ell,dy=-d\ell,dx=d\ell,dy=d\ell$, respectively (with the other differential being zero). Plugging those in fixes the apparent sign discrepancies in the first two integrals.

To deal with the general question, since I'm not really comfortable with differential forms, I like to think of $\int P dx + Q dy$ as being by definition $\int (P,Q) \cdot T d \ell$, where $T$ is the unit tangent. So let us start from the RHS of the first equality and rewrite it like this.

Notice that $n=AT$ where $A$ is the operator that performs clockwise rotation by 90 degrees. (Here we assume the curve is traversed counterclockwise and $n$ is the outward normal.) So now $v \cdot n = v \cdot (AT) = (A^* v) \cdot T$, where $A^*$ is a counterclockwise rotation by 90 degrees. And now $A^*(P,Q)=(Q,-P)$.

That's the slightly slick way to do it using some background in linear algebra; let me know if you'd rather I write it in the self-contained way using just coordinates.