The theorem
Before you start: you'll need line integrals and double integrals.
If \(C\) is a simple closed curve bounding a region \(R\), traversed anticlockwise, and \(M(x, y)\), \(N(x, y)\) have continuous partial derivatives, then
\[\oint_C \big(M\,dx + N\,dy\big) = \iint_R\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right)dx\,dy\]
Area as a line integral
\[A = \frac{1}{2}\oint_C \big(x\,dy - y\,dx\big)\]
The idea in one picture
The integrand \(\dfrac{\partial N}{\partial x} - \dfrac{\partial M}{\partial y}\) is the \(\mathbf{k}\)-component of the curl of \(M\mathbf{i} + N\mathbf{j}\). So Green's theorem says: the total swirl inside \(R\) equals the circulation round its edge. It is the flat (2D) case of Stokes' theorem.
Solved examples
Example 1: Verifying the theorem
Exam levelVerify Green's theorem for \(\displaystyle\oint_C \big[(xy + y^2)\,dx + x^2\,dy\big]\), where \(C\) bounds the region between \(y = x\) and \(y = x^2\).
- Double integral: \(N_x - M_y = 2x - (x + 2y) = x - 2y\). \(\displaystyle\int_0^1\!\!\int_{x^2}^{x}(x - 2y)\,dy\,dx = \int_0^1 (x^4 - x^3)\,dx = \frac{1}{5} - \frac{1}{4} = -\frac{1}{20}\).
- Along \(y = x^2\), \(x: 0 \to 1\): \(dy = 2x\,dx\). \(\displaystyle\int_0^1\big[(x^3 + x^4) + 2x^3\big]dx = \frac{3}{4} + \frac{1}{5} = \frac{19}{20}\).
- Along \(y = x\), \(x: 1 \to 0\): \(\displaystyle\int_1^0 3x^2\,dx = -1\).
- Total line integral: \(\dfrac{19}{20} - 1 = -\dfrac{1}{20}\). The two sides agree.
AnswerBoth sides equal \(-\dfrac{1}{20}\).
Example 2: Over a triangle
MediumUse Green's theorem to evaluate \(\displaystyle\oint_C \big[(3x - 8y^2)\,dx + (4y - 6xy)\,dy\big]\), where \(C\) is the triangle with vertices \((0, 0), (1, 0), (0, 1)\).
- \(N_x - M_y = -6y - (-16y) = 10y\).
- \(\displaystyle\iint_R 10y\,dA = 10\int_0^1\!\!\int_0^{1-x} y\,dy\,dx = 10\int_0^1\frac{(1 - x)^2}{2}\,dx = 10\cdot\frac{1}{6}\).
Answer\(\dfrac{5}{3}\)
Example 3: Area of an ellipse
MediumUse Green's theorem to find the area of the ellipse \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\).
- \(x = a\cos t\), \(y = b\sin t\). \(x\,dy - y\,dx = (ab\cos^2 t + ab\sin^2 t)\,dt = ab\,dt\).
- \(A = \dfrac{1}{2}\displaystyle\int_0^{2\pi} ab\,dt\).
Answer\(\pi ab\)
Example 4: A zero circulation
EasyShow that \(\displaystyle\oint_C \big(e^x\sin y\,dx + e^x\cos y\,dy\big) = 0\) for every closed curve \(C\).
- \(N_x = e^x\cos y\) and \(M_y = e^x\cos y\), so \(N_x - M_y = 0\) and the double integral is 0.
Answer\(0\). The field is \(\nabla(e^x\sin y)\), which is conservative.
Green's theorem turns a line integral round a complicated boundary (often three or four pieces) into one double integral. Unless the question says “verify”, use the double integral side; it's almost always shorter.
Common mistakes
1. Getting the order wrong in \(N_x - M_y\)
It is \(\dfrac{\partial N}{\partial x} - \dfrac{\partial M}{\partial y}\), where \(N\) goes with \(dy\). Swapping gives the wrong sign.
2. Going clockwise
The theorem assumes anticlockwise travel. Clockwise changes the sign.
3. Using it when \(M\) or \(N\) is undefined inside \(R\)
For example \(\dfrac{-y\,dx + x\,dy}{x^2 + y^2}\) round a circle about the origin: the theorem doesn't apply, because the functions blow up at the origin.
Practice questions
Q1\(\displaystyle\oint_C (y\,dx - x\,dy)\) round the circle of radius 2, anticlockwise
\(N_x - M_y = -1 - 1 = -2\). Answer: \(-2\cdot 4\pi = -8\pi\).
Q2\(\displaystyle\oint_C \big[(x^2 - y^2)\,dx + 2xy\,dy\big]\) round the square \(0 \le x, y \le 1\)
\(N_x - M_y = 2y + 2y = 4y\). \(\displaystyle\int_0^1\!\!\int_0^1 4y\,dy\,dx = 2\).
Q3Find the area of a circle of radius \(a\) using \(\tfrac{1}{2}\oint (x\,dy - y\,dx)\).
\(x = a\cos t\), \(y = a\sin t\): \(\tfrac{1}{2}\displaystyle\int_0^{2\pi}a^2\,dt = \pi a^2\).
Q4\(\displaystyle\oint_C \big[(2x - y)\,dx + (x + y)\,dy\big]\) round the unit circle
\(N_x - M_y = 1 + 1 = 2\). Answer: \(2\pi\).
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Frequently asked questions
What does Green's theorem say?
The circulation of a field round a closed curve equals the double integral of \(N_x - M_y\) over the region inside.
How is Green's theorem related to Stokes' theorem?
Green's theorem is Stokes' theorem for a flat surface lying in the \(xy\)-plane.
Where Green's theorem leads next
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Vipul Sir has taught mathematics for over 15 years to HSC, CBSE, ICSE, IGCSE/IB, Diploma, Engineering and BSc students at VVS Classes in Goregaon and Vile Parle, Mumbai. He teaches every class himself, concepts first and exam technique second.