# 16.4: Green's Theorem

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We now come to the first of three important theorems that extend the Fundamental Theorem of Calculus to higher dimensions. (The Fundamental Theorem of Line Integrals has already done this in one way, but in that case we were still dealing with an essentially one-dimensional integral.) They all share with the Fundamental Theorem the following rather vague description: *To compute a certain sort of integral over a region, we may do a computation on the boundary of the region that involves one fewer integrations.*

Note that this does indeed describe the Fundamental Theorem of Calculus and the Fundamental Theorem of Line Integrals: to compute a single integral over an interval, we do a computation on the boundary (the endpoints) that involves one fewer integrations, namely, no integrations at all.

If the vector field \({\bf F}=\langle P,Q\rangle\) and the region \(D\) are sufficiently nice, and if \(C\) is the boundary of \(D\) (\(C\) is a closed curve), then

$$\iint\limits_{D} {\partial Q\over\partial x}-{\partial P\over\partial y} \,dA = \int_C P\,dx +Q\,dy ,$$

provided the integration on the right is done counter-clockwise around \(C\).

To indicate that an integral \(\int_C\) is being done over a closed curve in the counter-clockwise direction, we usually write \(\oint _C\). We also use the notation \(\partial D\) to mean the boundary of \(D\) {\dfont oriented\/}\index{oriented curve} in the counterclockwise direction. With this notation, \(\oint_C=\int_{\partial D}\).

We already know one case, not particularly interesting, in which this theorem is true: If \(\bf F\) is conservative, we know that the integral \(\oint_C {\bf F}\cdot d{\bf r}=0\), because any integral of a conservative vector field around a closed curve is zero. We also know in this case that \(\partial P/\partial y=\partial Q/\partial x\), so the double integral in the theorem is simply the integral of the zero function, namely, 0. So in the case that \(\bf F\) is conservative, the theorem says simply that \(0=0\).

Example \(\PageIndex{1}\)

We illustrate the theorem by computing both sides of

\[\int_{\partial D} x^4 \, dx + xy \, dy= \iint\limits_{D} y - 0 \, dA,\]

where \(D\) is the triangular region with corners \((0,0)\), \((1,0)\), \((0,1)\).

Starting with the double integral:

$$\iint\limits_{D} y-0\,dA=\int_0^1\int_0^{1-x} y\,dy\,dx= \int_0^1 {(1-x)^2\over2}\,dx=\left.-{(1-x)^3\over6}\right|_0^1={1\over6}.$$

There is no single formula to describe the boundary of \(D\), so to compute the left side directly we need to compute three separate integrals corresponding to the three sides of the triangle, and each of these integrals we break into two integrals, the "\(dx\)'' part and the "\(dy\)'' part. The three sides are described by \(y=0\), \(y=1-x\), and \(x=0\). The integrals are then

$$\eqalign{ \int_{\partial D}\!\!\! x^4\,dx + xy\,dy&= \int_0^1 x^4\,dx+\int_0^0 0\,dy+\int_1^0 x^4\,dx+\int_0^1 (1-y)y\,dy+ \int_0^0 0\,dx+\int_1^0 0\,dy\cr &={1\over5}+0-{1\over5}+{1\over6}+0+0={1\over6}.\cr}

$$

Alternately, we could describe the three sides in vector form as \(\langle t,0\rangle\), \(\langle 1-t,t\rangle\), and \(\langle 0,1-t\rangle\). Note that in each case, as \(t\) ranges from 0 to 1, we follow the corresponding side in the correct direction. Now

$$\eqalign{ \int_{\partial D} x^4\,dx + xy\,dy&= \int_0^1 t^4 + t\cdot 0\,dt + \int_0^1 -(1-t)^4 + (1-t)t\,dt +\int_0^1 0 + 0\,dt\cr &=\int_0^1 t^4\,dt + \int_0^1 -(1-t)^4 + (1-t)t\,dt ={1\over6}.\cr }$$

In this case, none of the integrations are difficult, but the second approach is somewhat tedious because of the necessity to set up three different integrals. In different circumstances, either of the integrals, the single or the double, might be easier to compute. Sometimes it is worthwhile to turn a single integral into the corresponding double integral, sometimes exactly the opposite approach is best.

Here is a clever use of Green's Theorem: We know that areas can be computed using double integrals, namely,

$$\iint\limits_{D} 1\,dA$$

computes the area of region \(D\). If we can find \(P\) and \(Q\) so that \(\partial Q/\partial x-\partial P/\partial y=1\), then the area is also

$$\int_{\partial D} P\,dx+Q\,dy.$$

It is quite easy to do this: \(P=0,Q=x\) works, as do \(P=-y, Q=0\) and \(P=-y/2,Q=x/2\).

Example \(\PageIndex{2}\)

An ellipse centered at the origin, with its two principal axes aligned with the \(x\) and \(y\) axes, is given by

$${x^2\over a^2}+{y^2\over b^2}=1.$$

We find the area of the interior of the ellipse via Green's theorem. To do this we need a vector equation for the boundary; one such equation is \(\langle a\cos t,b\sin t\rangle\), as

\(t\) ranges from 0 to \(2\pi\). We can easily verify this by substitution:

$${x^2\over a^2}+{y^2\over b^2}={a^2\cos^2 t\over a^2}+{b^2\sin^2t\over b^2}= \cos^2t+\sin^2t=1.$$

Let's consider the three possibilities for \(P\) and \(Q\) above: Using 0 and \(x\) gives

$$\oint_C 0\,dx+x\,dy=\int_0^{2\pi} a\cos(t)b\cos(t)\,dt= \int_0^{2\pi} ab\cos^2(t)\,dt.$$

Using \(-y\) and 0 gives

$$\oint_C -y\,dx+0\,dy=\int_0^{2\pi} -b\sin(t)(-a\sin(t))\,dt= \int_0^{2\pi} ab\sin^2(t)\,dt.$$

Finally, using \(-y/2\) and \(x/2\) gives

$$\eqalign{\oint_C -{y\over2}\,dx+{x\over2}\,dy&= \int_0^{2\pi} -{b\sin(t)\over2}(-a\sin(t))\,dt +{a\cos(t)\over2}(b\cos(t))\,dt\cr &=\int_0^{2\pi} {ab\sin^2t\over2}+{ab\cos^2t\over2}\,dt=\int_0^{2\pi} {ab\over2}\,dt=\pi ab.\cr}$$

The first two integrals are not particularly difficult, but the third is very easy, though the choice of \(P\) and \(Q\) seems more complicated.

**Figure 16.4.1.** A "standard'' ellipse, \({x^2\over a^2}+{y^2\over b^2}=1\).

Proof of Green's Theorem

We cannot here prove Green's Theorem in general, but we can do a special case. We seek to prove that

$$\oint_C P\,dx +Q\,dy = \iint\limits_{D} {\partial Q\over\partial x}-{\partial P\over\partial y} \,dA.$$

It is sufficient to show that

$$\oint_C P\,dx=\iint\limits_{D}-{\partial P\over\partial y} \,dA\qquad\hbox{and}

\qquad\oint_C Q\,dy=\iint\limits_{D} {\partial Q\over\partial x}\,dA,$$

which we can do if we can compute the double integral in both possible ways, that is, using \(dA=dy\,dx\) and \(dA=dx\,dy\).

For the first equation, we start with

$$\iint\limits_{D}{\partial P\over\partial y}\,dA= \int_a^b\int_{g_1(x)}^{g_2(x)} {\partial P\over \partial y}\,dy\,dx= \int_a^b P(x,g_2(x))-P(x,g_1(x))\,dx.$$

Here we have simply used the ordinary Fundamental Theorem of Calculus, since for the inner integral we are integrating a derivative with respect to \(y\): an antiderivative of \(\partial P/\partial y\) with respect to \(y\) is simply \(P(x,y)\), and then we substitute \(g_1\) and \(g_2\) for \(y\) and subtract.

Now we need to manipulate \(\oint_C P\,dx\). The boundary of region \(D\) consists of 4 parts, given by the equations \(y=g_1(x)\), \(x=b\),\(y=g_2(x)\), and \(x=a\). On the portions \(x=b\) and \(x=a\), \(dx=0\,dt\), so the corresponding integrals are zero. For the other two portions, we use the parametric forms \(x=t\), \(y=g_1(t)\), \(a\le t\le b\), and \(x=t\), \(y=g_2(t)\), letting \(t\) range from \(b\) to \(a\), since we are integrating counter-clockwise around the boundary. The resulting integrals give us

$$\eqalign{

\oint_C P\,dx = \int_a^b P(t,g_1(t))\,dt+\int_b^a P(t,g_2(t))\,dt &=\int_a^b P(t,g_1(t))\,dt-\int_a^b P(t,g_2(t))\,dt\cr &=\int_a^b P(t,g_1(t))-P(t,g_2(t))\,dt\cr }$$

which is the result of the double integral times \(-1\), as desired.

The equation involving \(Q\) is essentially the same, and left as an exercise.

\( \square \)

## Contributors

Integrated by Justin Marshall.