Green's Theorem
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 6.4: “Green’s Theorem” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/6-4-greens-theorem |
| Textbook used in class | Stewart, Calculus, Section 16.4: “Green’s Theorem” (Theorem, Examples 1, 2) |
Quick Reference
Green’s Theorem: Let $D$ be a simply-connected region in $\mathbb{R}^2$ with positively-oriented (counterclockwise) boundary $C = \partial D$. If $P$ and $Q$ have continuous partial derivatives on $D$, then: $$\oint_C P\,dx + Q\,dy = \iint_D\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\,dA.$$
Positively oriented: the region $D$ is on the LEFT as you traverse $C$.
Motivation
Green’s theorem converts a line integral around a closed curve into a double integral over the enclosed region. This is useful in both directions: sometimes the double integral is easier than the line integral (simplifying computation), and sometimes the line integral is easier than the double integral (computing area via the boundary).
Green’s theorem is also a 2D version of Stokes’ theorem -- it is the bridge between the single-curve world and the surface world.
Key Concepts
1. Orientation
The positive orientation means counterclockwise traversal of the outer boundary. If $D$ has holes (multiply-connected), the outer boundary is traversed counterclockwise and the inner boundary (around each hole) clockwise.
2. The Integrand $Q_x - P_y$
The quantity $Q_x - P_y$ is related to the $z$-component of $\text{curl}\,\mathbf{F}$ (where $\mathbf{F} = \langle P, Q, 0\rangle$). For a conservative field, $Q_x = P_y$, so $Q_x - P_y = 0$, and Green’s theorem gives $\oint_C\mathbf{F}\cdot d\mathbf{r} = 0$ -- consistent with path independence.
Worked Example
Evaluate $\oint_C(y - x^2)\,dx + (x + y^3)\,dy$ where $C$ is the counterclockwise boundary of the triangle with vertices $(0,0)$, $(1,0)$, $(0,1)$. (Stewart 16.4, Example 1.)
$P = y-x^2$, $Q = x+y^3$.
$Q_x = 1$, $P_y = 1$.
$Q_x - P_y = 0$.
By Green’s theorem: $\oint_C\cdots = \iint_D 0\,dA = 0$.
(This field is conservative, so the line integral over any closed curve is zero.)
Example 2. Evaluate $\oint_C(-y^3\,dx + x^3\,dy)$ where $C$ is the unit circle (counterclockwise).
$P = -y^3$, $Q = x^3$.
$Q_x - P_y = 3x^2 - (-3y^2) = 3(x^2+y^2) = 3r^2$ (in polar).
$\oint_C\cdots = \iint_D 3(x^2+y^2)\,dA = \int_0^{2\pi}\int_0^1 3r^2\cdot r\,dr\,d\theta = 2\pi\int_0^1 3r^3\,dr = 2\pi\cdot\frac{3}{4} = \frac{3\pi}{2}.$
Green’s theorem applies to any closed curve, not just simple ones. Green’s theorem requires $C$ to be a simple closed curve (no self-intersections) and $D$ to be its interior. For self-intersecting curves or curves bounding a region with holes, modifications are needed. The “positive orientation” condition (region on the left) is also essential for the formula to hold with the correct sign.
Common Misconceptions
Green’s theorem holds for any closed curve traversed in any direction.
This is the concept-image-conflicts-definition error. Green’s theorem requires the boundary curve $C$ to be traversed with positive orientation: the enclosed region $D$ must lie to the left as $C$ is traversed, which means counterclockwise for a simple outer boundary. If $C$ is traversed clockwise, the theorem gives $\oint_C P\,dx + Q\,dy = -\iint_D(Q_x - P_y)\,dA$ -- the result has the opposite sign. Reversing orientation is a common error that flips every sign in the answer.
Leveled Practice
Problem 1. Use Green’s theorem to evaluate $\oint_C(x^2+y)\,dx + (y^2-x)\,dy$ where $C$ is the counterclockwise unit circle.
Show answer
$P = x^2+y$, $Q = y^2-x$.
$Q_x - P_y = -1 - 1 = -2$.
$\oint_C\cdots = \iint_D(-2)\,dA = -2\cdot\pi(1)^2 = -2\pi$.