Integral theorems · Definition
Green's Theorem
Before you start you need to know
It turns a line integral around a closed curve into a double integral over the region it encloses, and back; it is the planar version of Stokes' theorem.
Definition
Green's theorem equates the circulation of a field \(\mathbf F=\langle P,Q\rangle\) along a simple closed curve \(C\), counterclockwise, with a double integral over its region \(D\), with respect to area \(A\).
The hypotheses matter: \(C\) must be closed, simple and piecewise smooth, traversed with the region \(D\) always on the left (counterclockwise), and \(P\), \(Q\) must have continuous partial derivatives on an open set containing \(D\). Traversing \(C\) backwards flips the sign. If \(D\) has holes, the full boundary includes the inner curves traversed clockwise.
The integrand \(Q_x-P_y\) is the \(\mathbf k\)-component of the curl of \(\langle P,Q,0\rangle\): the total circulation along the boundary is the sum of the small local circulations inside \(D\), because contributions along shared interior edges cancel in pairs. A guiding example is \(\langle -y,x\rangle\), with \(Q_x-P_y=2\): its circulation around any simple counterclockwise curve is twice the enclosed area.
This yields area formulas, such as \(A=\oint_C x\,dy=-\oint_C y\,dx=\tfrac12\oint_C x\,dy-y\,dx\), and a quick way to evaluate hard line integrals: if \(Q_x-P_y\) is simple, integrate over the region instead. Typical mistakes: forgetting the orientation, or applying the theorem to an open curve, where it does not hold.
If \(Q_x=P_y\) on a region without holes, the circulation around every closed curve is zero, which is the underlying reason a field with zero curl is conservative there. The theorem generalizes in two directions: Stokes' theorem, to surfaces in space, and the divergence theorem, to flux across a boundary.
Think about it
What is \(\oint_C -y\,dx+x\,dy\) over a simple counterclockwise curve enclosing area \(7\)?
\(14\): since \(Q_x-P_y=2\), the integral is twice the area.
Why does Green's theorem not apply as is to a curve enclosing a hole where the field is undefined?
Because it requires \(P\) and \(Q\) to be differentiable throughout the enclosed region; if they fail at an interior point, the double integral makes no sense there.
Exercises 10
- Let \(C\) be the circle \(x^2+y^2=4\) traversed counterclockwise. Compute \(\oint_C -y\,dx+x\,dy\).
- Let \(C\) be the boundary of the rectangle \([0,3]\times[0,2]\), traversed counterclockwise. Compute \(\oint_C y\,dx\).
- Compute \(\oint_C xy\,dx+x^2\,dy\), where \(C\) is the counterclockwise boundary of the square \([0,1]\times[0,1]\).
- Compute \(\oint_C x^2\,dx+xy\,dy\), where \(C\) is the counterclockwise boundary of the triangle with vertices \((0,0)\), \((1,0)\) and \((0,1)\).
- Compute \(\oint_C y^3\,dx+x^3\,dy\) over the unit circle, counterclockwise.
- Let \(C\) be the ellipse \(\tfrac{x^2}{4}+\tfrac{y^2}{9}=1\), counterclockwise. Compute \(\oint_C -\tfrac y2\,dx+\tfrac x2\,dy\).
- If \(\oint_C P\,dx+Q\,dy=5\) with \(C\) counterclockwise, what is the integral over the same curve traversed clockwise?
- Let \(C\) be any simple closed piecewise smooth curve. What is \(\oint_C 2xy\,dx+x^2\,dy\)?
- Let \(C\) be the counterclockwise boundary of a region \(D\). Which of these line integrals over \(C\) gives the area of \(D\)?
- Compute \(\oint_C x\,dy\) over the counterclockwise boundary of the triangle with vertices \((0,0)\), \((4,0)\) and \((0,3)\).
What comes next
Related
References
- James Stewart, Daniel K. Clegg, Saleem Watson, Calculus: Early Transcendentals, 9th edition, Cengage, 2021. §16.4
- Joel Hass, Christopher Heil, Przemyslaw Bogacki, Maurice D. Weir, George B. Thomas Jr., Thomas' Calculus, 15th edition, Pearson, 2023. §16.4
- Jerrold E. Marsden, Anthony Tromba, Vector Calculus, 6th edition, W. H. Freeman, 2012. §8.1