Green's theorem

Green’s theorem converts a closed-curve Line integral in the plane into a double integral over the enclosed region. It comes in two equivalent forms — circulation and flux — corresponding to the Curl and Divergence of a 2D vector field.

The pattern matches every other rung of the FTC family: integral of derivative on the inside = boundary integral on the outside.

Setup

Let $D$ be a bounded planar region with a piecewise-smooth simple closed boundary $C=\partial D$, traversed in the positive (counterclockwise) direction so that $D$ is on the left. Let $\mathbf{F}(x,y)=⟨P(x,y),Q(x,y)⟩$ be a vector field with continuous first partials on a neighborhood of $\overline{D}$.

Circulation form

${\oint }_{C}Pdx+Qdy={\iint }_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA.$

The left side is the circulation $\oint \mathbf{F}\cdot d\mathbf{r}$ in component form. The right side is a double integral of $Q_x-P_y$ — which is the $\hat{\mathbf{k}}$-component of the 3D curl of $\mathbf{F}$ (viewing the 2D field as a 3D field with no $z$-component).

In vector form:

${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\iint }_D(\nabla \times \mathbf{F})\cdot \hat{\mathbf{k}}dA.$

Flux form

${\oint }_C\mathbf{F}\cdot \hat{\mathbf{N}}ds={\iint }_D\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}\right)dA={\iint }_D\nabla \cdot \mathbf{F}dA.$

The left side is the outward flux of $\mathbf{F}$ across $C$. The right side integrates divergence over $D$. “What flows out through the boundary = what’s being produced inside.”

Why they’re equivalent

Apply the circulation form to the rotated field $\mathbf{G}=⟨-Q,P⟩$. The left side becomes $\oint -Qdx+Pdy$, which equals the flux of $\mathbf{F}$. The right-side integrand becomes $P_x-(-Q{)}_y=P_x+Q_y=\nabla \cdot \mathbf{F}$.

Why Green’s is true (sketch)

For a rectangle $D=[a,b]\times [c,d]$, traversed counterclockwise:

The $\oint Pdx$ piece picks up contributions only from the horizontal sides (bottom $y=c$ left-to-right, top $y=d$ right-to-left), since $dx=0$ on the vertical sides. Combining:

$\oint Pdx={\int }_a^{b}P(x,c)dx-{\int }_a^{b}P(x,d)dx=-{\int }_a^b\left[P(x,d)-P(x,c)\right]dx.$

By the single-variable FTC applied to $P$ as a function of $y$,

$P(x,d)-P(x,c)={\int }_c^d\frac{\partial P}{\partial y}(x,y)dy.$

Substituting:

$\oint Pdx=-{\iint }_D\frac{\partial P}{\partial y}dA.$

The analogous argument with $Q$ on vertical sides gives $\oint Qdy={\iint }_D{Q}_{x}dA$. Adding the two yields Green’s circulation form.

For more general regions, decompose into rectangular strips and apply the rectangle case piecewise. The conceptual heart: Green’s = FTC applied twice and stitched together.

What Green’s does NOT require

$\mathbf{F}$ does not have to be conservative. This is crucial. If $\mathbf{F}$ happens to be conservative, $Q_x-P_y=0$ and Green’s just says $\oint =0$, which we already knew. The interesting cases are when $\mathbf{F}$ is not conservative.

Worked example 1

${\oint }_{C}2xydx-x^{2}dy$ over $C=$ the positively oriented circle $|z|=2$.

By Green: $P=2xy$, $Q=-x^2$. $Q_x=-2x$, $P_y=2x$, so $Q_x-P_y=-4x$. Integral becomes

${\iint }_D-4xdA$

over the disk $x^2+y^2\le 4$. In polar: ${\int }_0^{2\pi }{\int }_0^2(-4r\cos \theta )rdrd\theta =0$ (since ${\int }_0^{2\pi }\cos \theta d\theta =0$). Answer: $0$.

Worked example 2: computing area

Setting $Q_x-P_y=1$, Green’s gives $\oint Pdx+Qdy=\mathrm{Area}(D)$. Standard choices:

• $P=0,Q=x$: $\mathrm{Area}=\oint xdy$.
• $P=-y,Q=0$: $\mathrm{Area}=-\oint ydx$.
• Symmetric: $P=-y/2$, $Q=x/2$: $\mathrm{Area}=\frac{1}{2}\oint (xdy-ydx)$.

In polar coordinates the symmetric form becomes $\mathrm{Area}=\frac{1}{2}\int r^{2}d\theta$.

Multiply connected regions

For an annular region (one hole) — boundary is the outer curve counterclockwise plus the inner curve clockwise — Green’s reads

${\oint }_{\text{out}}\mathbf{F}\cdot d\mathbf{r}-{\oint }_{\text{in}}\mathbf{F}\cdot d\mathbf{r}={\iint }_D(Q_x-P_y)dA,$

with both circles oriented counterclockwise in the formula. The minus sign is the bookkeeping.

This resolves the 2D rotational field puzzle (see Conservative vector field): on an annulus avoiding the origin, $Q_x=P_y$, so Green’s gives the difference of circulations is zero — every closed curve enclosing the origin has the same circulation $2\pi$. This is the deformation principle of 2D vector calculus, and is the direct ancestor of the Deformation principle for contour integrals in complex analysis.

Special case of Stokes’

Green’s theorem is exactly Stokes’ theorem applied to a flat 2D region in the $xy$-plane (with $\hat{\mathbf{N}}=\hat{\mathbf{k}}$). The 3D theorem reduces to the 2D theorem when the surface is flat.

In the FTC family

Theorem	Geometry
FTC	0D bdry, 1D interior
FTLI	0D bdry, 1D curve
Green’s (both)	1D bdry, 2D plane
Stokes’ theorem	1D bdry, 2D surface in 3D
Divergence theorem	2D bdry, 3D volume