Green's Theorem

Green’s Theorem is an integral theorem that relates a line integral (integral over a parametric curve) to the double integral over the curve’s enclosed region. It is a special case of Stokes’ theorem.

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Definition §

Green's Theorem ^theorem

Let $D$ be a simply connected 2D region bounded by a smooth (potentially piecewise) simple, closed curve $C$ that is positively orientated. Then, let $P(x,y)$ and $Q(x,y)$ be $C^1$ functions that have a domain which $D$ is a subset oof. Green’s Theorem states:

$${\int }_C(P\mathrm{d}x+Q\mathrm{d}y)\mathrm{d}s={\iint }_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\mathrm{d}x\mathrm{d}y$$

• The left hand side is a path integral over $C$
• The right hand side is a double integral over $D$

Green's Theorem (Vector Form)

Let $D$ be a simply connected region in the X-Y plane bounded by a smooth (potentially piecewise) simple, closed curve $C$ that is positively orientated. Then, let $\vec{F}=(P(x,y),Q(x,y),0)$ be a $C^1$ vector field. Green’s Theorem states:

$${\int }_C\vec{F}\cdot \mathrm{d}\vec{c}={\iint }_D(\nabla \times \vec{F})\cdot \hat{k}\mathrm{d}x\mathrm{d}y$$

Derivation §

#tosee Proof for all cases

This derivation assumes that $D$ is both vertically simple and horizontally simple. We will show the right hand side is equal to the left side.

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Consider:

$${\iint }_D\frac{\partial P}{\partial y}\mathrm{d}y\mathrm{d}x$$

Since we assume $D$ is vertically simple, we can use the vertical strips decomposition. That is:

$$D=\{(x,y)\in {\mathbb{R}}^2:a\le x\le b,l(x)\le y\le u(x)\}$$

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$${\iint }_D\frac{\partial P}{\partial y}\text{d}x\text{d}y={\int }_a^b{\int }_{l(x)}^{u(x)}\frac{\partial P}{\partial y}\text{d}y\text{d}x$$

Cancelling the $y$-dependency:

$${\int }_a^b{\int }_{l(x)}^{u(x)}\frac{\partial P}{\cancel{\partial y}}\cancel{\text{d}y}\text{d}x={\int }_a^{b}P(x,u(x))-P(x,l(x))\text{d}x$$

The difference can be expressed as the decomposition of two curves, $C_1$ and $C_2$. Then we can express our integral as the difference of curve/path integrals

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We can treat $x$ as a parameter (like $t$). Then:

$${\int }_a^{b}P(x,u(x))-P(x,l(x))\text{d}x=\left(-{\int }_{C_1}P\text{d}x\right)-{\int }_{C_2}P\text{d}x$$

However, note that our orientation of $C$ means $C_2$ is orientated negatively (if we choose $t\in [u,v]$ such that from $c_2(u)=a$ and $c_2(v)=b$) therefore, it would be negative:

A curve integral can be split up, and also combined, if its over the same curve and function.

$$=-\left({\int }_{C_1}P\text{d}x+{\int }_{C_2}P\text{d}x\right)=-{\int }_{C}P\text{d}x$$

So then, we can take the negative of this:

$${\iint }_D-\frac{\partial P}{\partial y}\text{d}x\text{d}y={\int }_{C}P\text{d}x$$

Similarly, we can compute ${\iint }_D\frac{\partial Q}{\partial x}\text{d}x\text{d}y$:

This time, treat $D$ as horizontally simple:

$$D=\{(x,y)\in {\mathbb{R}}^2:c\le y\le d,l(y)\le x\le r(y)\}$$

Then:

$${\iint }_D\frac{\partial Q}{\partial x}\text{d}x\text{d}y={\int }_c^d{\int }_{l(y)}^{r(y)}\frac{\partial Q}{\cancel{\partial x}}\cancel{\text{d}x}\text{d}y={\int }_c^{d}Q(r(y),y)-Q(l(y),y)\text{d}y$$

Again, using a curve decomposition, we split $C$ into $C_3$ and $C_4$. Treating $y$ as a parameter, and remembering that $C_4$ is oriented negatively.

$${\int }_c^{d}Q(r(y),y)-Q(l(y),y)\text{d}y={\int }_{C_3}Q\text{d}y-\left(-{\int }_{C_4}Q\text{d}y\right)={\int }_{C_3}Q\text{d}y+{\int }_{C_4}Q\text{d}y={\int }_{C}Q\text{d}y$$

So finally, we have:

$${\int }_C(P\mathrm{d}x+Q\mathrm{d}y)\mathrm{d}s={\iint }_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\mathrm{d}x\mathrm{d}y$$