Math Core

The Fundamental Theorem for Line Integrals handles conservative fields. Green's theorem handles every field in the plane, as long as the curve is closed: it converts a line integral around a closed curve into a double integral over the region inside. Often one side is much easier than the other, and the theorem also gives a neat way to compute areas.

Orientation and the statement

A simple closed curve is a closed curve that doesn't cross itself, like a circle or the outline of a triangle. It is positively oriented if you traverse it counterclockwise, so the enclosed region $D$ is always on your left. The notation $\oint_C$ signals that $C$ is closed.

Green's theorem

Let $C$ be a positively oriented, piecewise-smooth, simple closed curve in the plane, and let $D$ be the region it encloses. If $P$ and $Q$ have continuous partial derivatives on an open region containing $D$ , then

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

The integrand $Q_x - P_y$ measures how much $\mathbf{F} = \langle P, Q\rangle$ circulates around each point. Green's theorem says the circulation around the boundary is the total of all the little circulations inside: neighboring little loops cancel along their shared edges, and only the outer boundary survives.

A special case is worth noticing. If $Q_x = P_y$ everywhere in $D$ , the right side is $0$ , which matches what you learned for conservative fields.

Worked example: A square

Evaluate $\displaystyle\oint_C xy\,dx + x^2\,dy$ , where $C$ is the boundary of the square $0 \le x \le 1$ , $0 \le y \le 1$ , oriented counterclockwise.

Doing this directly would take four line integrals, one per side. With Green's theorem, $P = xy$ and $Q = x^2$ , so $Q_x - P_y = 2x - x = x$ :

\oint_C xy\,dx + x^2\,dy = \int_0^1\!\!\int_0^1 x\,dy\,dx = \frac{1}{2}.

Worked example: When the line integral is hopeless

Evaluate $\displaystyle\oint_C \left(\sin(x^2) - y\right)dx + \left(3x + e^{y^2}\right)dy$ , where $C$ is the circle $x^2 + y^2 = 4$ , counterclockwise.

The functions $\sin(x^2)$ and $e^{y^2}$ have no elementary antiderivatives, so a direct parametrization gets stuck. But $Q_x - P_y = 3 - (-1) = 4$ , a constant. So

\oint_C = \iint_D 4\,dA = 4\cdot\text{area}(D) = 4\cdot 4\pi = 16\pi.

The triangle with vertices (0, 0), (2, 0) and (0, 2). Traversed C₁ → C₂ → C₃, it is positively oriented.Open in grapher →

Worked example: A triangle

Evaluate $\displaystyle\oint_C y^2\,dx + 3xy\,dy$ , where $C$ is the triangle with vertices $(0, 0)$ , $(2, 0)$ , $(0, 2)$ , oriented counterclockwise.

$Q_x - P_y = 3y - 2y = y$ . The region is $0 \le y \le 2$ , $0 \le x \le 2 - y$ :

\iint_D y\,dA = \int_0^2 y(2 - y)\,dy = 4 - \frac{8}{3} = \frac{4}{3}.

Computing area with a line integral

Run Green's theorem backwards. If you choose $P$ and $Q$ with $Q_x - P_y = 1$ , the double integral is just the area of $D$ . Three convenient choices:

Area formulas

For a positively oriented simple closed curve $C$ enclosing $D$ ,

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

Worked example: Area of an ellipse

Find the area enclosed by $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ .

Parametrize: $x = a\cos t$ , $y = b\sin t$ , $0 \le t \le 2\pi$ . Then $dx = -a\sin t\,dt$ and $dy = b\cos t\,dt$ , so

x\,dy - y\,dx = ab\cos^2 t\,dt + ab\sin^2 t\,dt = ab\,dt.

The area is $\dfrac{1}{2}\displaystyle\int_0^{2\pi} ab\,dt = \pi ab$ . With $a = b = r$ this is the familiar $\pi r^2$ .

Regions with holes

Green's theorem extends to regions with holes, such as an annulus. The boundary then has several pieces, and "positively oriented" still means the region stays on your left: the outer curve goes counterclockwise and each inner curve goes clockwise. Add the line integrals over all the pieces.

Common mistake

Green's theorem assumes counterclockwise orientation. If a problem traverses $C$ clockwise, the line integral is the negative of $\iint_D (Q_x - P_y)\,dA$ . Also check that $P$ and $Q$ are defined and smooth on all of $D$ ; a field that blows up inside $C$ needs the holes version.

Tip

The order in the integrand is " $Q_x$ minus $P_y$ ": the $x$ -derivative of the $dy$ coefficient minus the $y$ -derivative of the $dx$ coefficient.

Practice

Practice 1

Evaluate $\displaystyle\oint_C x\,dy$ , where $C$ is the boundary of the rectangle $0 \le x \le 3$ , $0 \le y \le 2$ , oriented counterclockwise.