[b][size=150]Green's Theorem[/size][/b][br][br]Suppose [math]R[/math] is a simply connected region i.e. no holes in [math]\mathbb{R}^2[/math] and the boundary of [math]R[/math] is a closed, piecewise smooth curve [math]C[/math] oriented counterclockwise. The following famous theorem states the connection between the line integral over [math]C[/math] and the double integral over [math]R[/math]:[br][br][u]Theorem[/u] [b](Green's theorem)[/b] Let [math]\vec{F}(x,y)=\langle f(x,y), g(x,y) \rangle[/math] be a 2D vector field such that both [math]f[/math] and [math]g[/math] are continuous and have continuous first order partial derivatives on some open region containing [math]R[/math]. Then [br][br][math]\oint_C\vec{F}\cdot d\vec{r}=\int_C f \ dx + g \ dy =\iint_R\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right) \ dA[/math][br]([u]Note[/u]: The symbol [math]\oint[/math] means [math]C[/math] is a closed curve.)[br][br][u]Proof[/u]: By definition, [math]\int_C\vec{F}\cdot d\vec{r}=\int_C f(x,y) \ dx+\int_C g(x,y) \ dy[/math].[br][br]To proof Green's theorem, it suffices to prove the following:[br][br](a) [math]\int_C f(x,y) \ dx=-\iint_R \frac{\partial f}{\partial y} \ dA[/math][br](b) [math]\int_C g(x,y) \ dy=\iint_R \frac{\partial g}{\partial x} \ dA[/math][br][br]Consider (a): [br][br]We regard [math]R[/math] as a type I region such that it is bounded from below by curve [math]C_1[/math], which is the graph of [math]y=g_1(x)[/math] and from above by [math]C_2[/math], which is the graph of [math]y=g_2(x)[/math] for [math]a\leq x \leq b[/math]. Then their parametrizations are [math]C_1:\vec{r}_1(t)=\langle t,g_1(t)\rangle[/math] and [math]-C_2:\vec{r}_2(t)=\langle t,g_2(t)\rangle[/math] for [math]a\leq t \leq b[/math]. [br][br][math]\int_C f(x,y) \ dx=\int_{C_1} f(x,y) \ dx -\int_{C_2} f(x,y) \ dx[/math][br][math]=\int_a^b f(t,g_1(t))\cdot 1 \ dt- \int_a^b f(t,g_2(t))\cdot 1 \ dt[/math][br][math]=\int_a^b (f(t,g_1(t))-f(t,g_2(t))) \ dt[/math][br][math]=-\int_a^b\left(\int_{g_1(t)}^{g_2(t)} \frac{\partial f}{\partial y} \ dy\right) \ dt[/math] (FTC)[br][math]=-\iint_R \frac{\partial f}{\partial y} \ dA[/math][br][br]Consider (b):[br][br]We regard [math]R[/math] as a type II region and use the similar argument, it can be shown that[br][br][math]\int_C g(x,y) \ dy=\iint_R \frac{\partial g}{\partial x} \ dA[/math][br][br]Combining (a) and (b), Green's theorem is proved.[br][br][br][u]Remark[/u]: If [math]\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}=0[/math], then [math]\vec{F}[/math] is conservative and [math]\oint_C\vec{F}\cdot d\vec{r}=0[/math].[br][br][br][br][u]Example[/u]: We revisit the earlier example - evaluate [math]\int_C x^2y \ dx + x \ dy[/math], where [math]C[/math] is the loop along the triangle from [math](0,0)[/math] to [math](1,0)[/math] to [math](1,2)[/math] and back to [math](0,0)[/math]. Here we use Green's theorem to find the line integral.[br][br]Let [math]f(x,y)=x^2y[/math] and [math]g(x,y)=x[/math]. Then we have[br][br][math]\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}=1-x^2[/math][br][br]By Green's theorem, [br][br][math]\int_C x^2y \ dx + x \ dy=\iint_R(1-x^2) \ dA[/math], where [math]R[/math] is the triangular region with vertices [math](0,0), (1,0)[/math] and [math](1,2)[/math].[br][br]Regarding [math]R[/math] as a type I region, we have[br][br][math]\iint_R(1-x^2) \ dA=\int_0^1\left(\int_0^{2x}(1-x^2) \ dy \right) \ dx[/math][br][math]=\int_0^1\left[(1-x^2)y\right]_0^{2x} \ dx[/math][br][math]=\int_0^1(2x-2x^3) \ dx=\left[x^2-\frac{x^4}2\right]_0^1=\frac 12[/math][br][br][u]Remark[/u]: The double integral is easier to compute than the sum of three line integrals over three sides of the triangle.[br][br]

[b][size=150]Applications[/size][/b][br][br][u]Finding area of a region[/u][br][br]We can use Green's theorem to find the area of a simply connected region [math]R[/math] using the vector field [math]\vec{F}(x,y)=\langle -y, x \rangle[/math]. Suppose we are given the parametrization of the boundary of [math]R[/math] (oriented counterclockwise). Then[br][br][math]\int_C\vec{F}\cdot d\vec{r}=\iint_R (1-(-1)) \ dA= 2\iint_R 1 \ dA=2\times \text{Area of }R[/math][br][br][u]Remark[/u]: Other vector fields like [math]\vec{F}(x,y)=\langle y, 0 \rangle[/math] also works. The mechanism of a useful tool called [url=https://en.wikipedia.org/wiki/Planimeter]plainmeter[/url], which is used for measuring the area of an arbitrary shape by tracing its boundary, is based on Green's theorem.[br][br][br][u]Example[/u]: Let [math]C[/math] be an ellipse parametrized by [math]\vec{r}(t)=\langle a\cos(t), b\sin(t) \rangle[/math] for [math]0\leq t \leq 2\pi[/math]. Use Green's theorem to find the area of the region [math]R[/math] enclosed by [math]C[/math].[br][br][u]Answer[/u]:[br][br][math]R[/math] is obviously a simply connected region. Then by Green's theorem, we have[br][br][math]2\iint_R 1 \ dA=\int_C\vec{F}\cdot d\vec{r}=\int_C -y \ dx+ x \ dy=\int_0^{2\pi}[(-b\sin(t))(-a\sin(t))+(a\cos(t))(b\cos(t))] \ dt[/math][br][math]=\int_0^{2\pi}ab(\sin^2(t)+\cos^2(t)) \ dt=2\pi ab [/math][br][br]Therefore, the area of the ellipse is [math]\pi ab[/math].[br]

[u]Multiply connected regions[/u][br][br]Although Green's theorem can only be applied to simply connected regions, we can use the following trick to deal with regions that have "holes":[br][br]Let [math]D[/math] be an annulus, as shown in the image below

The boundary of the annulus consists of two circles - the outer one is oriented counterclockwise and the inner one is oriented clockwise. Then we make two "cuts" at a point [math]P[/math] and divide [math]D[/math] into two simply connected regions [math]D_1[/math] and [math]D_2[/math], enclosed by the curves as shown above. By Green's theorem, we have[br][br][math]\iint_D\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right) \ dA=\iint_{D_1}\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right) \ dA+\iint_{D_2}\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right) \ dA[/math][br][math]=\int_{P_1\cup P_2 \cup P_3 \cup P_4} f \ dx+ g \ dy + \int_{P_5\cup -P_2 \cup P_6 \cup -P_4} f \ dx+ g \ dy[/math][br][math]=\int_{P_1}f \ dx+ g \ dy + \int_{P_2}f \ dx+ g \ dy + \int_{P_3}f \ dx+ g \ dy + \int_{P_4}f \ dx+ g \ dy [/math][br][math]+\int_{P_5}f \ dx+ g \ dy+\int_{-P_2}f \ dx+ g \ dy+\int_{P_6}f \ dx+ g \ dy+\int_{-P_4}f \ dx+ g \ dy[/math][br][math]=\int_{P_1\cup P_6} f \ dx+ g \ dy + \int_{P_3\cup P_5} f \ dx+ g \ dy[/math][br][br]Notice that [math]P_1\cup P_6[/math] is the outer boundary of [math]D[/math] (counterclockwise) and [math]P_3\cup P_5[/math] is the boundary of the "hole" of [math]D[/math] (clockwise). [br][br]This trick can also be applied to any multiply-connected region i.e. a connected region [math]R[/math] with [math]k[/math] holes. Suppose [math]C_0[/math] is the outer boundary of [math]R[/math] (counterclockwise) and [math]C_i[/math] is the boundary of the [math]i^{\text{th}}[/math] hole (clockwise), [math]i=1,2,\ldots, k[/math]. Make the cuts as shown in the diagram below. Then the Green's theorem for [math]R[/math] is as follows:[br][br][math]\int_{C_0+C_1+\cdots+C_k}f \ dx+ g \ dy=\iint_R \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right) \ dA[/math][br]

[u]Example[/u]: Let [math]\vec{F}(x,y)=\langle f(x,y),g(x,y)\rangle=\left\langle -\frac{y}{x^2+y^2}, \frac{x}{x^2+y^2}\right\rangle[/math] for [math](x,y)\ne (0,0)[/math]. Evaluate [math]\oint_C\vec{F}\cdot d\vec{r}[/math], where [math]C[/math] is a piecewise smooth, counterclockwise simple closed curve if [br][br](a) [math]C[/math] does not enclose [math](0,0)[/math].[br](b) [math]C[/math] encloses [math](0,0)[/math].[br][br][br][u]Answer[/u]:[br][br]First, notice that [math]\frac{\partial f}{\partial y}=\frac{\partial g}{\partial x}=\frac{y^2-x^2}{(x^2+y^2)^2}[/math].[br][br](a) Suppose [math]C[/math] does not enclose [math](0,0)[/math]. [math]\vec{F}[/math] is defined on the region [math]R[/math] enclosed by [math]C[/math] and [math]R[/math] is simply connected. Therefore, by Green's theorem, we have[br][br][math]\oint_C\vec{F}\cdot d\vec{r}=\iint_R\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right) \ dA=0[/math].[br][br](b) Since [math]\vec{F}[/math] is not defined at [math](0,0)[/math] i.e. [math]\vec{F}[/math] is not continuous on the region enclosed by [math]C[/math], Green's theorem cannot be directly applied. We consider the multiply-connected region [math]R'[/math] bounded by [math]C[/math] and [math]C_a[/math], which is the circle centered at [math](0,0)[/math] with radius [math]a>0[/math] (for small enough [math]a[/math]) and oriented clockwise, as shown in the applet below. Using Green's theorem for multiply-connected regions, we have[br][br][math]0=\iint_{R'}\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right) \ dA=\oint_C\vec{F}\cdot d\vec{r}+\int_{C_a}\vec{F}\cdot d\vec{r}[/math][br][math]\implies \oint_C\vec{F}\cdot d\vec{r}=-\int_{C_a}\vec{F}\cdot d\vec{r}=\int_{-C_a}\vec{F}\cdot d\vec{r}[/math][br][br]Consider the parametrization of [math]-C_a[/math]: [math]\langle a\cos(t),a\sin(t)\rangle[/math] for [math]0\leq t \leq 2\pi[/math]. We have[br][br][math]\int_{-C_a}\vec{F}\cdot d\vec{r}=\int_0^{2\pi}\left\langle \frac{-a\sin(t)}{a^2},\frac{a\cos(t)}{a^2}\right\rangle\cdot\langle -a\sin(t),a\cos(t)\rangle \ dt=\int_0^{2\pi}1 \ dt=2\pi[/math][br][br]Hence, [math]\oint_C\vec{F}\cdot d\vec{r}=2\pi[/math] for any simple closed curve [math]C[/math] that encloses [math](0,0)[/math].