Green's Theorem

Introduction

Recall that ∫C\displaystyle\int_C is a path integral. Now, we define ∮C\displaystyle\oint_C as a path integral along a closed path (meaning that the path returns to where it started).

For instance, using the fundamental theorem: ∫C∇ϕ⋅r⃗ ′(t) dt=ϕ(B)−ϕ(A)∮C∇ϕ⋅r⃗ ′(t) dt=0\begin{aligned} \int_C \nabla \phi \cdot \vec{r}\ '(t)\ dt &= \phi(B) - \phi(A)\ \oint_C \nabla \phi \cdot \vec{r}\ '(t)\ dt &= 0 \end{aligned}

Green's Theorem relates a path integral on a closed curve to a double integral on the region inside the curve. First, though, we need a few simple definitions:

Green's Theorem

There are two forms of Green's Theorem. First, though, let CC be a simple, closed, piecewise-smooth curve, positively oriented, that encloses a connected and simply connected region RR in the plane. Assume that F⃗=⟨f,g⟩\vec{F}=⟨f,g⟩, where ff and gg have continuous first partial derivatives in RR.

  1. Circulation, or Curl, Form ∮CF⃗⋅r⃗ ′(t) dt=∮Cf dx +g dy=∬R∂g∂x−∂f∂y dA\oint_C \vec{F} \cdot \vec{r}\ '(t)\ dt = \oint_C f\ dx \ + g\ dy = \iint_R \dfrac{\partial g}{\partial x} - \dfrac{\partial f}{\partial y} \ dA

  2. Flux, or Divergence, Form ∮CF⃗⋅n^ ds=∮Cf dy −g dx=∬R∂f∂x+∂g∂y dA,\oint_C \vec{F} \cdot \hat{n}\ ds = \oint_C f\ dy \ - g\ dx = \iint_R \dfrac{\partial f}{\partial x} + \dfrac{\partial g}{\partial y} \ dA, where n^\hat{n} is the outward unit normal on the curve.

Note that if you set up the circulation/curl form with F⃗=⟨−g,f⟩\vec{F}=\langle -g,f \rangle, you get the flux/divergence form.

Examples

Use Green's Theorem to evaluate ∮Cxy dx +x2y3 dy,\oint_C xy\ dx\ + x^2y^3\ dy, where CC is the triangle with vertices (0,0)(0,0), (1,0)(1,0), and (1,2)(1,2) with positive orientation.

Solution

Since f=xyf=xy and g=x2y3g=x^2y^3, ∂f∂y=x\dfrac{\partial f}{\partial y} = x and ∂g∂x=2xy3\dfrac{\partial g}{\partial x} = 2xy^3, so

∮Cxy dx +x2y3 dy=∫01∫02x2xy3−x dydx=∫018x5−2x2 dx=43x6−23x3∣01=23\begin{aligned} \oint_C xy\ dx\ + x^2y^3\ dy &= \int_0^1 \int_0^{2x} 2xy^3-x\ dy dx\ &= \int_0^1 8x^5-2x^2\ dx\ &= \dfrac{4}{3}x^6-\dfrac{2}{3}x^3 \bigg|_0^1 = \ans{\dfrac{2}{3}} \end{aligned}

  1. Use Green's Theorem to evaluate ∮Cx3 dy −y3 dx\displaystyle\oint_C x^3\ dy\ - y^3\ dx, where CC is the positively oriented circle of radius 22 centered at the origin.

    2. ∫02π∫023r2⋅r drdθ=24π\displaystyle\int_0^{2\pi} \displaystyle\int_0^2 3r^2 \cdot r \ dr d\theta = 24\pi

  2. Use Green's Theorem to evaluate ∮C(4x3+sin⁡(y2)) dy −(4y3+cos⁡(x2)) dx\displaystyle\oint_C (4x^3+\sin (y^2))\ dy\ - (4y^3+\cos (x^2))\ dx, where CC is the boundary of R={(x,y) : x2+y2≤4}R = {(x,y)\ :\ x^2+y^2 \leq 4}.

    3. 12∫02π∫02r2⋅r dr dθ=96π12\displaystyle\int_0^{2\pi} \displaystyle\int_0^2 r^2 \cdot r\ dr\ d\theta = 96\pi

  3. Use Green's Theorem to evaluate ∮C(2x+ey2) dy −(4y2+ex2) dx\displaystyle\oint_C \left(2x+e^{y^2}\right)\ dy\ - \left(4y^2+e^{x^2}\right)\ dx, where CC is the boundary of the square with vertices at (0,0)(0,0), (1,0)(1,0), (1,1)(1,1), and (0,1)(0,1).

    4. 66

Consider the rotation vector field F⃗=⟨−y,x⟩\vec{F}=⟨−y,x⟩ on the unit disk R={(x,y) : x2+y2≤1}.R = { (x,y) \ :\ x^2+y^2 \leq 1}.

Find the circulation ∬R∂g∂x−∂f∂y dA\displaystyle\iint_R \dfrac{\partial g}{\partial x} - \dfrac{\partial f}{\partial y} \ dA of this field.

Solution

Since f=−yf=−y and g=xg=x, ∂f∂y=−1\dfrac{\partial f}{\partial y} = -1 and \dfrac{\partial g}{\partial x} = 1, so ∬R∂g∂x−∂f∂y dA=∬R2 dA.\iint_R \dfrac{\partial g}{\partial x} - \dfrac{\partial f}{\partial y} \ dA = \iint_R 2\ dA.

Because the region is circular, this integral will be easiest using polar coordinates: ∬R2 dA=∫02π∫012r dr dθ=2π\iint_R 2\ dA = \int_0^{2\pi} \int_0^1 2r\ dr\ d\theta = \ans{2\pi}

  1. Suppose F⃗=∇(x2+y2)\vec{F}=∇(\sqrt{x^2+y^2}) and RR is the half annulus R={(r,θ):1≤r≤3,0≤θ≤π}R={(r,θ) : 1≤r≤3,0≤θ≤π}. Calculate the circulation on RR.

    2. 00

Consider the outward flux ∮CF⃗⋅n^ ds=∬R∂f∂x+∂g∂y dA\oint_C \vec{F} \cdot \hat{n}\ ds = \iint_R \dfrac{\partial f}{\partial x} + \dfrac{\partial g}{\partial y}\ dA of the radial field F⃗=⟨x,y⟩\vec{F}=⟨x,y⟩ across the unit circle C={(x,y):x2+y2=1}C={(x,y) : x^2+y^2=1}.

Solution

Since f=xf=x and g=yg=y, ∂f∂x=1\dfrac{\partial f}{\partial x} = 1 and ∂g∂y=1\dfrac{\partial g}{\partial y} = 1, so ∮CF⃗⋅n^ ds=∬R2 dA.\oint_C \vec{F} \cdot \hat{n}\ ds = \iint_R 2\ dA.

Because the region is circular, this integral will be easiest using polar coordinates: ∬R2 dA=∫02π∫012r dr dθ=2π\iint_R 2\ dA = \int_0^{2\pi} \int_0^1 2r\ dr\ d\theta = \ans{2\pi}

  1. Suppose F⃗=∇(x2+y2)\vec{F}=∇(\sqrt{x^2+y^2}) and RR is the half annulus R={(r,θ):1≤r≤3,0≤θ≤π}R={(r,θ) : 1≤r≤3,0≤θ≤π}. Calculate the outward flux across the boundary of RR.

    2. ∫0π∫13(−r2r3+2r)r dr dθ=2π\displaystyle\int_0^{\pi} \displaystyle\int_1^3 \left(-\dfrac{r^2}{r^3}+\dfrac{2}{r}\right) r\ dr\ d\theta = 2\pi

Consider the vector field F⃗=⟨y2,x2⟩\vec{F}=⟨y^2,x^2⟩ on the half annulus R={(x,y) : 1≤x2+y2≤9,y≥0}.R = {(x,y)\ : \ 1 \leq x^2+y^2 \leq 9, y \geq 0}.

Find the circulation on CC, the boundary of RR.

Solution

Recall that the circulation is ∮CF⃗⋅r⃗ ′(t) dt=∬R∂g∂x−∂f∂y dA\displaystyle\oint_C \vec{F} \cdot \vec{r}\ '(t)\ dt = \displaystyle\iint_R \dfrac{\partial g}{\partial x} - \dfrac{\partial f}{\partial y} \ dA.

Since f=y2f=y^2 and g=x2g=x^2, ∂f∂x=2y\dfrac{\partial f}{\partial x} = 2y and ∂g∂y=2x\dfrac{\partial g}{\partial y} = 2x, so ∮CF⃗⋅r⃗ ′(t) dt=∬R2x−2y dA.\oint_C \vec{F} \cdot \vec{r}\ '(t)\ dt = \iint_R 2x-2y\ dA.

Again, this integral will be easiest using polar coordinates: ∬R2x−2y dA=∫0π∫13(2r cos⁡θ−2r sin⁡θ)r dr dθ=−1043\iint_R 2x-2y\ dA = \int_0^{\pi} \int_1^3 (2r \ \cos\theta - 2r\ \sin\theta) r \ dr\ d\theta = \ans{-\dfrac{104}{3}}

Sidenote: using Green's Theorem to find area

If F⃗=⟨0,x⟩\vec{F}=⟨0,x⟩, ∮CF⃗⋅r⃗ ′(t) dt=∮Cx dy=∬R dA= area of R\oint_C \vec{F} \cdot \vec{r}\ '(t) \ dt = \oint_C x\ dy = \iint_R \ dA = \textrm{ area of } R

On the other hand, if F⃗=⟨y,0⟩\vec{F}=⟨y,0⟩, ∮CF⃗⋅r⃗ ′(t) dt=∮Cy dx=−∬R dA=−area of R\oint_C \vec{F} \cdot \vec{r}\ '(t) \ dt = \oint_C y\ dx = -\iint_R \ dA = -\textrm{area of } R

Therefore, the area of RR is ∮Cx dy=−∮Cy dx,\oint_C x\ dy = -\oint_C y\ dx, so Area of R=12∮Cx dy−y dx.\textrm{Area of } R = \dfrac{1}{2} \oint_C x\ dy - y\ dx.

Find the area of R={(x,y):x2+y2≤16}R={(x,y) : x^2+y^2≤16}.

Solution

The boundary of this region can be expressed parametrically as

x=4cos⁡t⟶dx=−4sin⁡t dty=4sin⁡t⟶dy=4cos⁡t dt\begin{aligned} x &= 4 \cos t \longrightarrow dx = -4 \sin t\ dt\ y &= 4 \sin t \longrightarrow dy = 4 \cos t\ dt \end{aligned}

The area of this region, then, is 12∮Cx dy−y dx=12∮C(4cos⁡t)(4cos⁡t)−(4sin⁡t)(4sin⁡t) dt=8∮Ccos⁡2t+sin⁡2t dt=8∫02π dt=16π\begin{aligned} &\dfrac{1}{2} \oint_C x\ dy - y\ dx\ = &\dfrac{1}{2} \oint_C (4 \cos t)(4 \cos t) - (4 \sin t)(4 \sin t)\ dt\ = &8 \oint_C \cos^2 t + \sin^2 t\ dt\ = &8 \int_0^{2\pi} \ dt = \ans{16\pi} \end{aligned}

  1. Use Green's Theorem to find the area of a disk of radius 55.

    2. 25π25\pi