6.2 Green's Theorem in the Plane

Green's Theorem bridges line integrals and double integrals, letting you convert a difficult line integral around a closed curve into a (usually easier) double integral over the region it encloses. It's one of the central results in multivariable calculus and shows up constantly in physics and engineering when you need to compute work, flux, or enclosed areas.

Green's Theorem Fundamentals

Application of Green's Theorem

Green's Theorem says that if you have a line integral around a simple, closed, counterclockwise curve $C$, you can trade it for a double integral over the region $D$ that $C$ encloses:

$\oint_C (P\, dx + Q\, dy) = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$

For this to work, a few conditions must hold:

• $C$ must be a simple closed curve (no self-intersections) traversed counterclockwise
• $D$ is the region enclosed by $C$
• $P$ and $Q$ must have continuous first partial derivatives on an open region containing $D$

That last condition is easy to overlook. If $P$ or $Q$ has a discontinuity or undefined partial derivative inside $D$, you can't apply the theorem directly.

Steps to apply Green's Theorem:

1. Identify the closed curve $C$ and the enclosed region $D$
2. Read off $P$ and $Q$ from the line integral ($P$ is the coefficient of $dx$, $Q$ is the coefficient of $dy$)
3. Compute the partial derivatives $\frac{\partial Q}{\partial x}$ and $\frac{\partial P}{\partial y}$
4. Set up and evaluate the double integral $\iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$

The whole point is that the double integral is often much simpler to evaluate than parameterizing the curve and computing the line integral directly.

Area Calculation with Green's Theorem

A particularly useful special case: you can compute the area of a region using a line integral. If you choose $P = -y$ and $Q = x$, then $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - (-1) = 2$, and Green's Theorem gives:

$\text{Area of } D = \frac{1}{2} \oint_C (x\, dy - y\, dx)$

There are two other equivalent forms that also work: $\oint_C x\, dy$ and $-\oint_C y\, dx$. All three give the same area; the symmetric version above is the most common.

Steps for area calculation:

1. Parameterize the closed curve $C$ (e.g., for an ellipse: $x = a\cos t$, $y = b\sin t$, $0 \le t \le 2\pi$)

2. Express $dx$ and $dy$ in terms of your parameter

3. Substitute into the area formula $\frac{1}{2} \oint_C (x\, dy - y\, dx)$

4. Evaluate the integral

For example, applying this to the ellipse parameterization above gives area $= \pi ab$, which matches the known formula.

Orientation of Closed Curves

Orientation matters. Green's Theorem as stated assumes counterclockwise (positive) orientation. If your curve runs clockwise, the result picks up a negative sign, so you'd need to negate the integral or reverse the curve's direction.

How to determine orientation:

• Parametric method: Track the direction of travel as the parameter increases. If the enclosed region stays to your left, you're going counterclockwise.
• Right-hand rule: Point your right thumb along the positive $z$-axis. Your fingers curl in the counterclockwise direction.

Regions with holes require extra care. If $D$ has a hole, the boundary of $D$ consists of multiple curves: the outer boundary traversed counterclockwise and each inner boundary (around a hole) traversed clockwise. Both orientations are chosen so that the region $D$ always lies to the left of the direction of travel. Getting this wrong flips signs in your answer.

Work and Flux Using Green's Theorem

Green's Theorem converts both work and flux line integrals into double integrals, but the setup differs between the two.

Work (circulation form):

For a force field $\mathbf{F}(x,y) = P\,\mathbf{i} + Q\,\mathbf{j}$, the work done along a closed curve is:

$W = \oint_C \mathbf{F} \cdot d\mathbf{r} = \oint_C (P\, dx + Q\, dy) = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$

This is the standard form of Green's Theorem. The integrand $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ is sometimes called the scalar curl of $\mathbf{F}$ (it's the $z$-component of $\nabla \times \mathbf{F}$).

Flux (divergence form):

The outward flux of $\mathbf{F}$ across $C$ uses a different line integral:

$\text{Flux} = \oint_C \mathbf{F} \cdot \mathbf{n}\, ds = \oint_C (P\, dy - Q\, dx) = \iint_D \left(\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}\right) dA$

Notice the integrand here is $\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}$, which is the divergence of $\mathbf{F}$. Don't mix up the two forms: circulation uses the curl, flux uses the divergence.

Problem-solving steps:

1. Determine whether you're computing work (circulation) or flux

2. Identify $P$ and $Q$ from the vector field

3. For work: compute $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$. For flux: compute $\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}$

4. Set up the double integral over $D$ with the appropriate integrand

5. Evaluate using whatever coordinate system fits the region (Cartesian, polar, etc.)

A quick sanity check: if the scalar curl is zero everywhere in $D$, the work around any closed curve in that region is zero (the field is conservative). If the divergence is zero everywhere, the net flux out of any closed curve is zero (the field is incompressible).