Applications of 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” (Corollary, Examples 3, 4) |
Quick Reference
Area formulas via Green’s theorem: $$\text{Area}(D) = \oint_C x\,dy = -\oint_C y\,dx = \frac{1}{2}\oint_C(x\,dy - y\,dx).$$
Regions with holes: apply Green’s theorem to each piece; the hole’s boundary is traversed clockwise (opposite to the region’s outer boundary).
Motivation
Green’s theorem provides a remarkable consequence: you can compute the area of a region purely by integrating around its boundary. This is the basis for planimeters (mechanical area-measuring devices) and digital surveying algorithms (the shoelace formula). It also gives a practical method for evaluating complicated double integrals by converting them to simpler line integrals.
Key Concepts
1. Area from the Boundary
Setting $Q = x$ and $P = 0$ in Green’s theorem: $Q_x - P_y = 1$, so: $$\text{Area}(D) = \iint_D 1\,dA = \oint_C x\,dy.$$
Setting $P = -y$ and $Q = 0$: $Q_x - P_y = 1$, so: $$\text{Area}(D) = \oint_C(-y)\,(-dx) = -\oint_C y\,dx.$$
Averaging: Area $= \frac{1}{2}\oint_C(x\,dy - y\,dx)$.
2. Regions with Holes
If $D$ has a hole with inner boundary $C_2$ (enclosing the hole), Green’s theorem gives: $$\oint_{C_1(\text{CCW})} + \oint_{C_2(\text{CW})} = \iint_D(Q_x-P_y)\,dA.$$
Worked Example
Find the area enclosed by the ellipse $x = a\cos t$, $y = b\sin t$, $0 \leq t \leq 2\pi$. (Stewart 16.4, Corollary.)
Using $\text{Area} = \frac{1}{2}\oint(x\,dy - y\,dx)$:
$dx = -a\sin t\,dt$, $dy = b\cos t\,dt$.
$$\text{Area} = \frac{1}{2}\int_0^{2\pi}(a\cos t\cdot b\cos t - b\sin t\cdot(-a\sin t))\,dt = \frac{1}{2}\int_0^{2\pi}(ab\cos^2 t + ab\sin^2 t)\,dt = \frac{ab}{2}\cdot 2\pi = \pi ab.$$
This confirms the standard formula: area of an ellipse with semi-axes $a$ and $b$ is $\pi ab$.
the area formula $A = \oint x\,dy$ requires counterclockwise orientation. Yes, the formula gives the area with a positive sign for counterclockwise orientation and a negative sign for clockwise. If you traverse the boundary clockwise, $\oint x\,dy = -\text{Area}(D)$. Always check the orientation before interpreting the sign of the result.
Common Misconceptions
the area formula $A = \oint_C x\,dy$ gives the area only when the region is convex.
This is the concept-image-conflicts-definition error. The formula $A = \oint_C x\,dy$ (and the equivalent forms $-\oint_C y\,dx$ and $\frac{1}{2}\oint_C(x\,dy - y\,dx)$) follow directly from Green’s theorem and apply to any region $D$ bounded by a simple closed curve with positive (counterclockwise) orientation, convex or not. For example, the formula correctly gives the area of a non-convex star-shaped region, provided the boundary is traced counterclockwise and encloses a simply-connected domain.
Leveled Practice
Problem 1. Find the area enclosed by the astroid $x = \cos^3 t$, $y = \sin^3 t$, $0 \leq t \leq 2\pi$.
Show answer
$A = \frac{1}{2}\int_0^{2\pi}(x\,dy/dt - y\,dx/dt)\,dt$.
$dx/dt = -3\cos^2 t\sin t$, $dy/dt = 3\sin^2 t\cos t$.
$x(dy/dt) - y(dx/dt) = \cos^3 t\cdot 3\sin^2 t\cos t - \sin^3 t\cdot(-3\cos^2 t\sin t) = 3\cos^4 t\sin^2 t + 3\cos^2 t\sin^4 t = 3\cos^2 t\sin^2 t$.
$A = \frac{3}{2}\int_0^{2\pi}\cos^2 t\sin^2 t\,dt = \frac{3}{2}\int_0^{2\pi}\frac{\sin^2 2t}{4}\,dt = \frac{3}{8}\cdot\pi = \frac{3\pi}{8}$.