Green’s theorem

A closed curve in the plane bounds a region, and every vector field defined on that region secretly links two very different-looking quantities: how much the field circulates around the boundary, and how much it "curls" over the interior. Green's theorem is the precise statement of that link — it converts a line integral around a closed curve into a double integral over the enclosed region, and back again. It sits in the MA10002 vector calculus unit right after line integrals, as the flat, two-dimensional special case of the family completed later by Stokes's theorem (3D circulation) and the divergence theorem (3D flux). You are expected to state it precisely, verify it on a concrete curve and field by computing both sides independently, and use it as a computational shortcut — swapping a line integral for a double integral, or a double integral for a line integral, whichever side is easier. It also hands you a free formula for the area enclosed by any curve.

The core insight

Green's theorem says the circulation of a vector field around a positively oriented closed curve equals the double integral of its scalar curl over the region the curve encloses, $\oint_C(M\,dx+N\,dy)=\iint_R(N_x-M_y)\,dA$ .

The core idea

Setup. Let $C$ be a positively oriented (counterclockwise), piecewise-smooth, simple closed curve in the plane, bounding a region $R$ . Let $M(x,y)$ and $N(x,y)$ have continuous partial derivatives on an open region containing $R$ .

The statement. Green's theorem asserts:

\oint_C (M\,dx + N\,dy) = \iint_R \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) dA.

where: $C$ is the boundary curve traversed counterclockwise, $R$ is the region it encloses, $M,N$ are the components of the vector field $\vec F=(M,N)$ , and $\partial N/\partial x - \partial M/\partial y$ is the scalar (out-of-plane) curl of $\vec F$ .

Reading it as circulation vs curl. The left side, $\displaystyle\oint_C\vec F\cdot d\vec r$ , is the total circulation of $\vec F$ around the boundary. The right side sums up the local rotation (curl) of $\vec F$ at every point of $R$ . Green's theorem says these two totals must agree — a purely local property (curl at each point) determines a global property (net circulation), which is why the theorem is such a powerful shortcut.

Two directions of use. If the double integral is easier, use it to avoid parametrizing $C$ piece by piece. If the line integral is easier — or the region $R$ is awkward to set up — go the other way and reduce a double integral to a boundary line integral instead.

The area formula. Choosing $M=-\tfrac12y,\ N=\tfrac12x$ makes $N_x-M_y=1$ , so the double integral becomes just the area of $R$ :

\text{Area}(R) = \frac12\oint_C (x\,dy - y\,dx).

where: the right side is a line integral computed purely from the boundary curve — no need to describe the interior of $R$ at all. Equivalent single-term choices $M=0,N=x$ or $M=-y,N=0$ give $\text{Area}(R)=\oint_C x\,dy=-\oint_C y\,dx$ .

Link to conservative fields. If $\vec F=(M,N)$ is curl-free, $N_x=M_y$ , then the right side of Green's theorem vanishes identically, so $\displaystyle\oint_C\vec F\cdot d\vec r=0$ for every closed curve in $R$ — exactly the zero-circulation property of conservative fields met earlier in the course, now derived as a special case.

Figure

Positively oriented boundary curve enclosing a region with the curl of the field distributed over the interior

Derivation AGreen's theorem as the 2D flux-of-curl identity

The goal is to show carefully, for a region $R$ that is both "vertically simple" (bounded above and below by graphs $y=g_2(x)\ge y=g_1(x)$ over $a\le x\le b$ ) and "horizontally simple," that the double integral $\iint_R(N_x-M_y)\,dA$ reduces exactly to the boundary line integral $\oint_C(M\,dx+N\,dy)$ , by handling the $M$ term and the $N$ term separately with the fundamental theorem of calculus and matching signs to the counterclockwise orientation.

Step 1. Handle the $M$ part first: show $\displaystyle\iint_R\left(-\frac{\partial M}{\partial y}\right)dA=\oint_C M\,dx$ . Treat $R$ as vertically simple, $a\le x\le b$ , $g_1(x)\le y\le g_2(x)$ . Then $\displaystyle\iint_R\left(-\frac{\partial M}{\partial y}\right)dA=-\int_a^b\int_{g_1(x)}^{g_2(x)}\frac{\partial M}{\partial y}\,dy\,dx=-\int_a^b\big[M(x,g_2(x))-M(x,g_1(x))\big]\,dx$ , using the fundamental theorem of calculus on the inner integral.

Step 2. Recognize the right side as a line integral. Split $C$ into the lower piece $C_1$ ( $y=g_1(x)$ , $x$ increasing from $a$ to $b$ ) and the upper piece $C_2$ ( $y=g_2(x)$ , $x$ decreasing from $b$ to $a$ , since counterclockwise traversal goes along the top backwards). Then $\displaystyle\int_a^b M(x,g_1(x))\,dx=\int_{C_1}M\,dx$ and $\displaystyle-\int_a^b M(x,g_2(x))\,dx=\int_{C_2}M\,dx$ (the minus sign appears because $C_2$ runs from $b$ to $a$ ). Adding these two pieces gives exactly $\displaystyle\oint_C M\,dx=-\iint_R\frac{\partial M}{\partial y}\,dA$ .

Step 3. Repeat symmetrically for the $N$ part, treating $R$ as horizontally simple ( $c\le y\le d$ , $h_1(y)\le x\le h_2(y)$ ), to obtain $\displaystyle\oint_C N\,dy=\iint_R\frac{\partial N}{\partial x}\,dA$ by the same fundamental-theorem argument along $x$ instead of $y$ .

Step 4. Add the two results from Step 2 and Step 3. $\displaystyle\oint_C M\,dx+\oint_C N\,dy=-\iint_R\frac{\partial M}{\partial y}\,dA+\iint_R\frac{\partial N}{\partial x}\,dA$ , which rearranges to $\displaystyle\oint_C(M\,dx+N\,dy)=\iint_R\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dA$ — precisely Green's theorem. General regions are handled by cutting them into finitely many pieces that are each both vertically and horizontally simple; the interior cuts cancel in pairs because each interior boundary is traversed once in each direction.

Worked example

Problem. Verify Green's theorem for $\vec F=(M,N)=(xy,\ x^2)$ on the region $R$ bounded by $y=x^2$ and $y=x$ between $x=0$ and $x=1$ , with $C$ the positively oriented boundary.

Step 1. Double-integral side. Compute $N_x-M_y=2x-x=x$ . On $R$ , for $0\le x\le1$ , $y$ runs from $y=x^2$ (lower) to $y=x$ (upper). So $\displaystyle\iint_R x\,dA=\int_0^1\int_{x^2}^{x}x\,dy\,dx=\int_0^1 x(x-x^2)\,dx=\int_0^1(x^2-x^3)\,dx=\left[\frac{x^3}3-\frac{x^4}4\right]_0^1=\frac13-\frac14=\frac1{12}$ .

Step 2. Line-integral side, piece $C_1$ : $y=x^2$ , $x:0\to1$ , $dy=2x\,dx$ . $\displaystyle\int_{C_1}(xy\,dx+x^2dy)=\int_0^1\big[x\cdot x^2+x^2\cdot2x\big]dx=\int_0^1 3x^3\,dx=\frac34$ .

Step 3. Piece $C_2$ : $y=x$ , traversed back from $(1,1)$ to $(0,0)$ , so $x:1\to0$ , $dy=dx$ . $\displaystyle\int_{C_2}(xy\,dx+x^2dy)=\int_1^0\big[x^2+x^2\big]dx=\int_1^0 2x^2\,dx=2\left[\frac{x^3}3\right]_1^0=-\frac23$ .

Step 4. Sum the two pieces: $\displaystyle\oint_C\vec F\cdot d\vec r=\frac34-\frac23=\frac9{12}-\frac8{12}=\frac1{12}$ . Both sides equal $\dfrac1{12}$ , confirming Green's theorem on this curve and field.

Competitive stretch

Applying the area formula to the ellipse $x^2/a^2+y^2/b^2=1$ , parametrized by $x=a\cos t,\ y=b\sin t$ for $t\in[0,2\pi]$ , gives $x\,dy-y\,dx=(a\cos t)(b\cos t\,dt)-(b\sin t)(-a\sin t\,dt)=ab\,dt$ , so $\text{Area}=\frac12\int_0^{2\pi}ab\,dt=\pi ab$ — the familiar ellipse area, obtained entirely from a boundary line integral, with no need to set up a double integral over the interior at all.

Common traps

Using the wrong orientation: Green's theorem requires $C$ positively oriented (counterclockwise, region on the left as you walk along $C$ ); a clockwise traversal flips the sign of the answer.
Forgetting that $M,N$ must be continuously differentiable throughout the closed region $R$ , not just on $C$ — a field with a singularity inside $R$ (such as $1/(x^2+y^2)$ at the origin) invalidates a direct application.
Sign slip in $N_x-M_y$ : it is $N_x$ minus $M_y$ , not $M_y$ minus $N_x$ ; reversing the order flips the sign of the whole double integral.
Setting up the double integral with the wrong bounds when the two curves bounding $R$ cross more than once — always sketch $R$ first to identify the correct upper and lower (or left and right) functions, as in the worked example above.

Ask the tutor: Want to see Green's theorem verified on a curve with three straight sides instead of a parabola and a line, or work through what changes when the region has a hole in it?

Why does Green's theorem need C to be positively oriented?

The proof builds the double integral by splitting $C$ into a lower piece traversed left-to-right and an upper piece traversed right-to-left (see the derivation above), and the signs from the fundamental theorem of calculus only line up correctly for that counterclockwise traversal. Reverse the direction and every sign in the derivation flips, so the line integral picks up an overall minus sign relative to the double integral. Practically: always check that you are walking along $C$ with the enclosed region on your left before applying the formula directly.

Can I use Green's theorem if the field has a singularity inside the region?

Not directly — the theorem requires $M$ and $N$ to have continuous partial derivatives on an open set containing the whole closed region $R$ , including the interior. If the field blows up at a point inside $R$ (a common example is a field like $1/(x^2+y^2)$ at the origin), you have to exclude a small disk around the singularity, apply Green's theorem on the resulting punctured region, and handle the small circle around the singularity separately. Simply plugging into the formula as if nothing were wrong gives a wrong answer.

What is the fastest way to decide whether to compute the line integral or the double integral?

Look at both sides before committing. If $N_x-M_y$ simplifies to something simple like a constant or a low-degree polynomial, and $R$ has clean bounds, the double integral is usually faster — this is exactly what happens in the ellipse area example. If instead the curve is simple to parametrize (a line, a circle, an ellipse) but the region's bounds would be awkward, or if $N_x-M_y$ is messy, compute the line integral directly. There is no universal rule; a quick glance at $N_x-M_y$ versus the shape of $C$ usually settles it in seconds.

Green’s theorem

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads