Green’s Theorem

The fundamental theorem of calculus and the fundamental theorem of line integrals share a common spirit: what happens in the interior of a region is expressed as an integral over its boundary. This can be viewed as the prototype of later results such as the divergence theorem and Stokes’ theorem, and Green’s theorem, which we examine in this post, can be regarded as the \(2\)-dimensional version of this spirit.

Green’s Theorem

First, for a planar region \(D\), among the two directions on its boundary curve \(\partial D\), we define the direction that keeps the region on the left (that is, traversing the outer boundary counterclockwise) as the positive orientation. Then the following holds.

Theorem 1 (Green) If \(D\) is a planar region bounded by a piecewise smooth simple closed curve \(C = \partial D\), and \(P, Q\), on an open set containing \(D\), are \(C^1\), then with \(C\) taken in the positive orientation,

\[\oint_C P\dd{x} + Q\dd{y} = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\dd{A}\]

holds.

Proof

First, in the case where \(D\) is simple with respect to \(y\), that is,

\[D = \{(x,y) \mid a \leq x \leq b,\ g_1(x) \leq y \leq g_2(x)\}\]

we show that the equation

\[\oint_C P\dd{x} = -\iint_D \partial P/\partial y\dd{A}\]

holds. For the double integral, evaluating the inner integral first using the iterated integral formula from §Multiple Integrals, §§Fubini’s Theorem,

\[\iint_D \frac{\partial P}{\partial y}\dd{A} = \int_a^b \bigl(P(x, g_2(x)) - P(x, g_1(x))\bigr)\dd{x}\]

holds. On the other hand, the boundary \(C\) consists of two pieces: traversing the lower curve \(y = g_1(x)\) with \(x\colon a \rightarrow b\) and the upper curve \(y = g_2(x)\) with \(x\colon b \rightarrow a\), while on the two vertical sides \(x\) is constant so \(\dd{x} = 0\). Therefore, since

\[\oint_C P\dd{x} = \int_a^b P(x, g_1(x))\dd{x} + \int_b^a P(x, g_2(x))\dd{x} = -\int_a^b \bigl(P(x,g_2) - P(x,g_1)\bigr)\dd{x}\]

comparing the two equations,

\[\oint_C P\dd{x} = -\iint_D \partial P/\partial y\dd{A}\]

holds. Symmetrically, if \(D\) is simple with respect to \(x\), then

\[\oint_C Q\dd{y} = \iint_D \partial Q/\partial x\dd{A}\]

holds. For a general region, cutting it into such pieces and summing them causes the integrals over the interior boundaries to cancel as two integrals with opposite directions, so the theorem holds.

In particular, if the integrand of the double integral is made \(1\) by choosing \(P, Q\), the area of the region can be computed as a line integral over the boundary.

Corollary 2 The area of \(D\) is given by the boundary integral:

\[\area(D) = \oint_C x\dd{y} = -\oint_C y\dd{x} = \frac{1}{2}\oint_C (x\dd{y} - y\dd{x})\]

Proof

In Theorem 1 (Green), if we set \((P, Q) = (0, x)\), then \(Q_x - P_y = 1\), so

\[\oint_C x\dd{y} = \iint_D 1\dd{A} = \area(D)\]

and if we set \((P, Q) = (-y, 0)\), then

\[\oint_C -y\dd{x} = \area(D)\]

holds. The third formula is the average of these two.

Meanwhile, Green’s theorem can be rewritten in two forms that interpret the two differential quantities of the planar vector field \(\mathbf{F} = (P, Q)\) as boundary integrals. Here, the divergence of a planar vector field, just as in the case of curl, is obtained by viewing \(\mathbf{F}\) as independent of \(z\) in the form \((P, Q, 0)\), which gives \(\divergence \mathbf{F} = \partial P/\partial x + \partial Q/\partial y\). (§Vector Fields, ⁋Definition 3)

Proposition 3 If \(D\) has a positively oriented boundary \(C\) and \(\mathbf{F} = (P, Q)\) is \(C^1\), then for the unit tangent \(\mathbf{T}\) and the outward unit normal \(\mathbf{n}\),

\[\oint_C \mathbf{F} \cdot \mathbf{T}\dd{s} = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\dd{A}, \qquad \oint_C \mathbf{F} \cdot \mathbf{n}\dd{s} = \iint_D \divergence \mathbf{F}\dd{A}\]

hold.

Proof

For the first identity,

\[\oint_C \mathbf{F}\cdot \mathbf{T}\dd{s} = \oint_C P\dd{x} + Q\dd{y}\]

is precisely the left-hand side of Theorem 1 (Green), and the integrand \(Q_x - P_y\) on the right-hand side is the curl of the planar vector field. (§Vector Fields, ⁋Definition 3)

For the second identity, using that on the positively oriented boundary the outward unit normal \(\mathbf{n}\) satisfies \(\mathbf{n}\dd{s} = (\dd{y}, -\dd{x})\), we have

\[\oint_C \mathbf{F}\cdot \mathbf{n}\dd{s} = \oint_C P\dd{y} - Q\dd{x}\]

and applying Theorem 1 (Green) to \((P, Q) \mapsto (-Q, P)\) shows that this equals

\[\iint_D (P_x + Q_y)\dd{A} = \iint_D \divergence \mathbf{F}\dd{A}\]

The first identity is exactly Green’s theorem, and only the second is new, but its intuitive meaning is also clear. That is, integrating the function \(\mathbf{F}\) along the boundary in the direction outward from the boundary is precisely what is contained in the divergence. On the other hand, we have already seen in §Line Integrals, ⁋Example 6 that even when curl vanishes, a vector field may fail to be conservative if the region has holes; this can be written rigorously as follows.

A region being simply connected means that any closed curve inside it can be continuously shrunk to a point without leaving the region; intuitively, one may think of this as a region without holes. For instance, a disk is simply connected, but a disk with its center removed is not, because to shrink a circle surrounding the center to a point one must necessarily pass through the missing center.

Corollary 4 In a simply connected open region, if a \(C^1\) vector field \(\mathbf{F} = (P, Q)\) satisfies \(\partial Q/\partial x = \partial P/\partial y\), then \(\mathbf{F}\) is conservative.

Proof

Since the region is simply connected, for any simple closed curve \(C\) inside it, the entire region \(D\) enclosed by it again lies within the region. Now

\[\oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_D (Q_x - P_y)\dd{A} = 0\]

and since the integral over every closed curve is \(0\), by §Line Integrals, ⁋Theorem 4, \(\mathbf{F}\) is conservative.

Example 5 (Simple connectedness) The vector field from §Line Integrals, ⁋Example 6,

\[\mathbf{F} = (-y, x)/(x^2+y^2)\]

was irrotational, but its integral over the unit circle around the origin was \(2\pi\). By Corollary 4, this only makes sense if the region on which this vector field is defined is not simply connected, and indeed the domain \(\mathbb{R}^2\setminus\{0\}\) is not simply connected. Moreover, over any closed curve not enclosing the origin the integrals of \(\mathbf{F}\) are all \(0\), so we can also confirm that the only problematic point is the origin where the vector field is undefined.

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References

[Ste] J. Stewart, Calculus, 8th ed., Cengage Learning, 2016.
[Kim] 김홍종, 미적분학 1·2, 제3개정판, 서울대학교출판문화원, 2020.