Green's Theorem

Let D be a "nice" region in the plane (where "'nice" means the boundary $\partial D$ has a continuous parametrization and does not intersect itself). The boundary should be traversed in the counterclockwise direction. Suppose that $\vec F = (P(x, y), Q(x, y))$ is a vector field, and P and Q have continuous partial derivatives. Green's theorem relates the line integral around the boundary $\partial D$ to the double integral over D:

$$\int_{\partial D} P(x, y)\,dx + Q(x, y)\,dy = \dint_D \left(\pder Q x - \pder P y\right)\,dx\,dy.$$

$$\hbox{\epsfysize=1.75 in \epsffile{greens-theorem-1.eps}}$$

Green's theorem is a special case of Stokes' theorem; to peek ahead a bit, $\pder Q x - \pder P y$ is just the z component of the $\curl$ of $\vec F$ , where $\vec F$ is regarded as a 3-dimensional vector field with zero z component:

$$\curl (P(x, y), Q(x, y), 0) = \left|\matrix{ \ihat & \jhat & \khat \cr \pder {} x & \pder {} y & \pder {} z \cr P(x, y) & Q(x, y) & 0 \cr}\right| = \left(0, 0, \pder Q x - \pder P y\right).$$

Example. Let D be the unit disk $x^2 + y^2 \le 1$ . Its boundary $\partial D$ is the unit circle $x^2 + y^2 = 1$ , which has the parametrization

$$x = \cos t, \quad y = \sin t, \quad 0 \le t \le 2 \pi.$$

$$\hbox{\epsfysize=1.75 in \epsffile{greens-theorem-2.eps}}$$

Verify that Green's theorem holds for the line integral

$$\int_{\partial D} y^3\,dx - x^3\,dy.$$

First, I'll compute the line integral directly.

$$\der x t = -\sin t, \quad \der y t = \cos t.$$

So

$$\int_{\partial D} y^3\,dx - x^3\,dy = \int_0^{2 \pi} \left((\sin t)^3(-\sin t) - (\cos t)^3(\cos t)\right)\,dt = \int_0^{2 \pi} \left(-(\cos t)^4 - (\sin t)^4\right)\,dt.$$

Apply the double angle formulas:

$$-(\cos t)^4 - (\sin t)^4 = -\dfrac{1}{4} \left((1 + \cos 2 t)^2 + (1 - \cos 2 t)^2\right) = -\dfrac{1}{4} \left(2 + 2 (\cos 2 t)^2\right) = -\dfrac{1}{2} \left(1 + \dfrac{1}{2}(1 + \cos 4 t)\right).$$

So I have

$$\int_0^{2 \pi} -\dfrac{1}{2} \left(1 + \dfrac{1}{2}(1 + \cos 4 t)\right)\,dt = -\dfrac{3\pi}{2}.$$

On the other hand, Green's theorem says the line integral is equal to

$$\dint_D \left(\pder Q x - \pder P y\right)\,dx\,dy = \dint_{x^2 + y^2 \le 1} \left(-3 x^2 - 3 y^2\right)\,dx\,dy.$$

Convert to polar. The region is

$$\left\{\matrix{ 0 \le r \le 1 \cr \noalign{\vskip2 pt} 0 \le \theta \le 2 \pi \cr}\right\}$$

So the integral becomes

$$\int_0^1 \int_0^{2 \pi} -3 r^2 \cdot r\,d\theta\,dr = -\dfrac{3 \pi}{2}.$$

The results agree.

Example. Verify Green's theorem for $\vec F = (x - y, x + y)$ and the curve $\vec\sigma(t) = (\cos t, \sin t)$ , $0 \le t \le 2 \pi$ .

The curve is the unit circle again, and the region D it encloses is the disk $x^2 + y^2 \le 1$ .

$$\der x t = -\sin t, \quad \der y t = \cos t.$$

So the line integral is

$$\int_{\vec\sigma} (x - y)\,dx + (x + y)\,dy = \int_0^{2 \pi} \left((\cos t - \sin t)(-\sin t) + (\cos t + \sin t) (\cos t)\right)\,dt = \int_0^{2 \pi}\,dt = 2 \pi.$$

Green's theorem says that the line integral is equal to the double integral

$$\dint_D \left(\pder Q x - \pder P y\right)\,dx\,dy = \dint_{x^2 + y^2 \le 1} \left(1 - (-1)\right)\,dx\,dy = 2\cdot (\hbox{the area of the circle}) = 2 \pi.$$

The results agree.

Example. Let R be the region bounded below by the x-axis, bounded on the right by $x = 1 - y$ for $0 \le y \le 1$ , and bounded on the left by $x = y - 1$ for $0 \le y \le 1$ . Compute

$$\int_{\partial R} (x^2 + y^2)\,dx + (y^2 + 8 x y)\,dy.$$

$$\hbox{\epsfysize=1.5 in \epsffile{greens-theorem-3.eps}}$$

The region is

$$\left\{\matrix{ 0 \le y \le 1 \cr y - 1 \le x \le 1 - y \cr}\right\}$$

$$\pder Q x - \pder P y = 8 y - 2 y = 6 y.$$

By Green's theorem,

$$\int_{\partial R} (x^2 + y^2)\,dx + (y^2 + 8 x y)\,dy = \int_0^1 \int_{y - 1}^{1 - y} 6 y\,dx\,dy = \int_0^1 6 y \left[x\right]_{y - 1}^{1 - y}\,dy = \int_0^1 6 y (2 - 2 y)\,dy =$$

$$\int_0^1 (12 y - 12 y^2)\,dy = \left[6 y^2 - 4 y^3\right]_0^1 = 2.\quad\halmos$$

Example. Let D be the region bounded by $y = \sin x$ , from $x = 0$ to $x = \pi$ , and the x-axis. Compute

$$\int_{\partial D} (x + y)^2\,dx - (x - y)^2\,dy,$$

Assume that the boundary is traversed counterclockwise.

$$\hbox{\epsfysize=1.5 in \epsffile{greens-theorem-4.eps}}$$

The region is

$$0 \le x \le \pi, \quad 0 \le y \le \sin x.$$

By Green's theorem,

$$\int_{\partial D} (x + y)^2\,dx - (x - y)^2\,dy = \int_0^\pi \int_0^{\sin x} \left(-2(x - y) - 2(x + y)\right)\,dy\,dx = \int_0^\pi \int_0^{\sin x} -4 x\,dy\,dx =$$

$$\int_0^\pi -4 x \sin x\,dx = -4 \pi.$$

If you compute the line integral directly, you need to parametrize the segment which makes up the base of the region and the curve. However, the curve is $y = \sin x$ as x goes from $\pi$ to 0, because the boundary of the region is traversed counterclockwise.

Example. Let P be the parallelogram with vertices $A(2, 1)$ , $B(4, 2)$ , $C(3, 4)$ , and $D(5, 5)$ . Compute

$$\int_{\partial P} (-2 y + 3 x^2 y + x y^2)\,dx + (x^2 y + x^3 + 3 x)\,dy,$$

Assume the boundary is traversed counterclockwise.

$$\hbox{\epsfysize=1.75 in \epsffile{greens-theorem-5.eps}}$$

To do the line integral directly, I'd need to parametrize each side and compute the integral over each side. Rather than do four integrals, I'll use Green's theorem:

$$\int_{\partial P} (-2 y + 3 x^2 y + x y^2)\,dx + (x^2 y + x^3 + 3 x)\,dy = \dint_P \left((2 x y + 3 x^2 + 3) - (-2 + 3 x^2 + 2 x y)\right)\,dx\,dy =$$

$$\dint_P 5\,dx\,dy = 5 \cdot (\hbox{the area of P}).$$

I don't need to compute the double integral; the area of a parallelogram is the length of the cross product of the vectors for two adjacent sides. $\bvec{A B} = (2, 1, 0)$ and $\bvec{A C} = \langle 1, 3, 0)$ , so

$$\bvec{A B} \times \bvec{A C} = \left|\matrix{ \ihat & \jhat & \khat \cr 2 & 1 & 0 \cr 1 & 3 & 0 \cr}\right| = (0, 0, 5).$$

Now $|\bvec{A B} \times \bvec{A C}| = 5$ , so the integral is $5 \cdot 5 = 25$ .

In the last two examples, the double integral reduced to a number times the area of the region. You can use Green's theorem to find the area of a region D as follows.

$$\int_{\partial D} x\,dy = \dint_D \left(1 - 0\right)\,dx\,dy = \hbox{the area of D}.$$

Example. The trisectrix of MacLaurin is given by the parametric equations

$$x = 1 - 4 (\cos t)^2, \quad y = (\tan t) \left(1 - 4 (\cos t)^2\right).$$

$$\hbox{\epsfysize=1.75 in \epsffile{greens-theorem-6.eps}}$$

(The trisectrix is the pedal curve of a parabola; the pedal point is the reflection of the focus across the directrix.)

The entire curve is traced out from $-\dfrac{\pi}{2}$ to $\dfrac{\pi}{2}$ . The loop is traced out from $-\dfrac{\pi}{3}$ to $\dfrac{\pi}{3}$ . Find the area of the region enclosed by the loop.

$$x \der y t = \left(1 - 4 (\cos t)^2\right) \left[(\sec t)^2 \left(1 - 4 (\cos t)^2\right) + (\tan t) (8) (\sin t) (\cos t)\right] =$$

$$\left(1 - 4 (\cos t)^2\right) \left((\sec t)^2 - 4 + 8 (\sin t)^2\right) = (\sec t)^2 + 8 (\cos t)^2 - 32 (\sin t)^2 (\cos t)^2.$$

The area is

$$\int_{-\pi/3}^{\pi/3} \left[(\sec t)^2 + 8 (\cos t)^2 - 32 (\sin t)^2 (\cos t)^2\right]\,dt = 3 \sqrt{3} = 5.19615 \ldots.$$

(You can integrate the second and third terms using the double angle formulas for sine and cosine.)

Example. (a) Parametrize $(x + y)^3 = 8 x y$ . (Hint: Let $y = x t$ .)

(b) Find the area enclosed by the loop of the curve.

(a) Following the hint, set $y = x t$ . Then

$$\eqalign{ (x + x t)^3 & = 8 x^2 t \cr x^3 (1 + t)^3 & = 8 x^2 t \cr x & = \dfrac{8 t}{(1 + t)^3} \cr}$$

Hence, $y = x t = \dfrac{8 t^2}{(1 + t)^3}$ .

$$\hbox{\epsfysize=1.75 in \epsffile{greens-theorem-7.eps}}$$

$y = x t$ is a line through the origin with slope t. Thus, the curve is being parametrized by the slope of the line joining the origin to a point on the curve.

(b) The loop is traced as t goes from 0 to $\infty$ .

$$\der y t = \dfrac{8 t(2 - t)}{(1 + t)^4}.$$

Hence,

$$x \der y t = \dfrac{8 t}{(1 + t)^3} \cdot \dfrac{8 t(2 - t)}{(1 + t)^4}.$$

The area is

$$\int_0^{\infty} \dfrac{8 t}{(1 + t)^3} \cdot \dfrac{8 t(2 - t)}{(1 + t)^4}\,dt.$$

You can do this integral by letting $u = 1 + t$ , so $du = dt$ . As t goes from 0 to $\infty$ , u goes from 1 to $\infty$ . The integral becomes

$$\int_1^{\infty} \dfrac{64 (u - 1)^2 (3 - u)}{u^7}\,du = 64 \lim_{b \to \infty} \int_1^b \dfrac{(u - 1)^2 (3 - u)}{u^7}\,du = 64 \lim_{b \to \infty} \left[\dfrac{1}{3 u^3} - \dfrac{5}{4 u^4} + \dfrac{7}{5 u^5} - \dfrac{1}{2 u^6}\right]_1^b =$$

$$64 \lim_{b \to \infty} \left(\dfrac{1}{3 b^3} - \dfrac{5}{4 b^4} + \dfrac{7}{5 b^5} - \dfrac{1}{2 b^6} + \dfrac{1}{60}\right) = \dfrac{16}{15}.$$

Here's how to do the antiderivative:

$$\int \dfrac{(u - 1)^2 (3 - u)}{u^7}\,du = \int \dfrac{3 - 7 u + 5 u^2 - u^3}{u^7}\,du =$$

$$\int \left(\dfrac{3}{u^7} - \dfrac{7}{u^6} + \dfrac{5}{u^5} - \dfrac{1}{u^4}\right)\,du = \dfrac{1}{3 u^3} - \dfrac{5}{4 u^4} + \dfrac{7}{5 u^5} - \dfrac{1}{2 u^6} + c.$$

(Multiply out the u stuff on top, then divide each term by $u^7$ .)

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