Green’s Theorem

2 MIN READ · PRACTISE AT YOUR OWN PACE · FREE

Boundary and interior

Green’s theorem connects circulation around a boundary to the sum of local rotation inside it. For a positively oriented, piecewise smooth boundary of a planar region $D$ , with $P,Q$ continuously differentiable on a neighborhood of the region,

\oint_{\partial D} P\,dx+Q\,dy=\iint_D(Q_x-P_y)\,dA.

Positive orientation keeps the region on your left: the outer boundary is counterclockwise, while boundaries of holes are clockwise.

Visual guide

VISUAL GUIDEThe interior remains on the left

The triangle with vertices (0, 0), (1, 0), (0, 1) is traversed counterclockwise. Green’s theorem replaces its boundary integral of x² dy by the integral of 2x over the shaded interior.

Choose the easier side

Draw the region first. Confirm that the field is defined throughout it, including the interior. A complicated boundary integral can simplify to an area calculation when $Q_x-P_y$ is constant. Conversely, a convenient boundary parametrization can evaluate an area integral.

Worked example: an ellipse

For $P=-y,Q=x$ , the scalar curl is $2$ . Around $x^2/a^2+y^2/b^2=1$ counterclockwise, circulation is $2\pi ab$ . The direct parametrization $(a\cos t,b\sin t)$ gives integrand $ab$ , agreeing with the theorem.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

In Green’s theorem, what is the positive orientation of a simple outer boundary in the plane?

Counterclockwise

Clockwise

Either direction with the same sign

Hint 1 · Find a starting point

Positive orientation keeps the enclosed region on your left.

Hint 2 · Take the next step

Walk around a circle and check which direction does this.

Show the reasoning

Answer: Counterclockwise

The standard formula uses counterclockwise orientation. Reversing it reverses the circulation’s sign.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: triangular region

Let $P=0,Q=x^2$ and $D=\{0\le x\le1,0\le y\le1-x\}$ . Then

\oint_{\partial D}x^2\,dy=\int_0^1 2x(1-x)\,dx=\frac13.

On the diagonal from $(1,0)$ to $(0,1)$ , set $x=1-t,y=t$ : its contribution is $\int_0^1(1-t)^2dt=1/3$ . The other two edges contribute zero.

Holes and exclusions

For a region with a hole, include both boundary components. You cannot fill in a singularity to make the integral simpler. Green’s flux form is $\oint(P\,dy-Q\,dx)=\iint(P_x+Q_y)\,dA$ for the same positive orientation; it measures outward planar flux.

Practice

Find circulation of $(-y,x)$ around the unit square counterclockwise.
Reverse the direction.
Find the outward flux of $(x,y)$ through the boundary of a disk of radius $2$ .

Show worked solutions

Curl $2$ times area $1$ gives $2$ .
Reversal gives $-2$ .
Divergence is $2$ and area is $4\pi$ , so outward flux is $8\pi$ . Flux and circulation use different derivatives.

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.

MAKE IT YOURS

Pause before the next idea.

Can you explain how the visualization connects to this lesson’s goal? If a step still feels uncertain, put this lesson on your review list and try the check again another day.

Optional marks, not a grade. Saved in this browser only. Open notebook →