Green's theorem

Say we have a piecewise smooth, closed curve

C

\mathbb R^2

bounding a region

R

\mathbb R^2

, oriented counter-clockwise; and a smooth vector field

\mathbf F=(P, Q)

also in

\mathbb R^2

Green’s theorem states that integrating a function along

C

is equal to integrating the curl of that function over the area

R

Normal form

Say we have the same vector field

\mathbf F

and curve

C

enclosing region

R

. We can also use Green’s theorem to relate the integral across

C

\mathbf F \cdot \hat{\mathbf{n} }

(the outwards-pointing vector normal to the curve) to an integral over

R

Proof

The proof for the normal form is given in terms of Green’s theorem.

We can write the integral as $\oint_C \mathbf F \cdot \hat{\mathbf n} ds$ . Note that $\hat{\mathbf n} ds$ has the same magnitude and is perpendicular to the differential change in position $d\mathbf r$ . If $d\mathbf r = (dx,dy)$ , then $\hat{\mathbf n} = (dy, -dx)$ .

The dot product is then $\mathbf F \cdot \hat{\mathbf n}ds = Pdy - Qdx$ .

Now let $\mathbf G = (-Q, P)$ . This means $\mathbf F \cdot \hat{\mathbf n}ds = \mathbf G \cdot d\mathbf r$ .

By Green’s theorem, $\oint_C \mathbf G \cdot d\mathbf r = \iint_R \mathrm{curl}(\mathbf G) dA$ .

$\mathrm{curl}(\mathbf G) = \mathrm{curl} \begin{pmatrix}-Q \\ P\end{pmatrix} = P_x + Q_y = \mathrm{div}(\mathbf F)$ .

Area

Green’s theorem can be used to find the area of a 2D shape. The integral we would normally use to calculate the area is just

\iint_R 1\,dA

. In order to use Green’s theorem, we need to find a function

\mathbf F = (P, Q)

such that

Q_x - P_y = \mathrm{curl}(\mathbf F) = 1

. By inspection we can find

\mathbf F = (0, x)

\mathbf F = (-y, 0)

Then we can integrate

\oint_C \mathbf F \cdot d\mathbf r

over the boundary of the region to find its area.