6.4 Green’s theorem  (Page 2/12)

this version of Green’s theorem is sometimes referred to as the tangential form of Green’s theorem .

The proof of Green’s theorem is rather technical, and beyond the scope of this text. Here we examine a proof of the theorem in the special case that D is a rectangle. For now, notice that we can quickly confirm that the theorem is true for the special case in which $\text{F}=⟨P,Q⟩$ is conservative. In this case,

because the circulation is zero in conservative vector fields. By [link] , F satisfies the cross-partial condition, so $P_y=Q_x.$ Therefore,

which confirms Green’s theorem in the case of conservative vector fields.

Proof

Let’s now prove that the circulation form of Green’s theorem is true when the region D is a rectangle. Let D be the rectangle $\left[a,b\right]\times \left[c,d\right]$ oriented counterclockwise. Then, the boundary C of D consists of four piecewise smooth pieces $C_1,$ $C_2,$ $C_3,$ and $C_4$ ( [link] ). We parameterize each side of D as follows:

Then,

By the Fundamental Theorem of Calculus,

Therefore,

But,

Therefore, ${\displaystyle {\int }_C\text{F}•d\text{r}={\displaystyle \int {\displaystyle {\int }_D\left(Q_x-P_y\right)dA}}}$ and we have proved Green’s theorem in the case of a rectangle.

To prove Green’s theorem over a general region D , we can decompose D into many tiny rectangles and use the proof that the theorem works over rectangles. The details are technical, however, and beyond the scope of this text.

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Applying green’s theorem over a rectangle

Calculate the line integral

${\displaystyle {\oint }_C{x}^{2}ydx+(y-3)dy},$

where C is a rectangle with vertices $(1,1),$ $(4,1),$ $(4,5),$ and $(1,5)$ oriented counterclockwise.

Let $\text{F}\left(x,y\right)=⟨P\left(x,y\right),Q\left(x,y\right)⟩=⟨x^{2}y,y-3⟩.$ Then, $Q_x=0$ and $P_y=x^2.$ Therefore, $Q_x-P_y=\text{−}{x}^2.$

Let D be the rectangular region enclosed by C ( [link] ). By Green’s theorem,

$\begin{array}{cc}\hfill {\displaystyle {\oint }_C{x}^{2}ydx+(y-3)dy}& ={\displaystyle {\iint }_D\left(Q_x-P_y\right)dA}\hfill \\ & ={\displaystyle \int {\displaystyle {\int }_D\text{−}{x}^{2}dA}}={\displaystyle {\int }_1^5{\displaystyle {\int }_1^4\text{−}{x}^{2}dxdy}}\hfill \\ & ={\displaystyle {\int }_1^5\mathrm{-21}dy}=\mathrm{-84.}\hfill \end{array}$

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Applying green’s theorem to calculate work

Calculate the work done on a particle by force field

$\text{F}(x,y)=⟨y+\text{sin}x,e^y-x⟩$

as the particle traverses circle $x^2+y^2=4$ exactly once in the counterclockwise direction, starting and ending at point $(2,0).$

Let C denote the circle and let D be the disk enclosed by C . The work done on the particle is

$W={\displaystyle {\oint }_C\left(y+\text{sin}x\right)dx+(e^y-x)dy}.$

As with [link] , this integral can be calculated using tools we have learned, but it is easier to use the double integral given by Green’s theorem ( [link] ).

Let $\text{F}\left(x,y\right)=⟨P\left(x,y\right),Q\left(x,y\right)⟩=⟨y+\text{sin}x,e^y-x⟩.$ Then, $Q_x=\mathrm{-1}$ and $P_y=1.$ Therefore, $Q_x-P_y=\mathrm{-2}.$

By Green’s theorem,

$\begin{array}{cc}\hfill W& ={\displaystyle {\oint }_C(y+\text{sin}(x))dx+(e^y-x)dy}\hfill \\ & ={\displaystyle {\iint }_D\left(Q_x-P_y\right)dA}={\displaystyle {\iint }_D\mathrm{-2}dA}\hfill \\ & =\mathrm{-2}\left(\text{area}\left(D\right)\right)=\mathrm{-2}\pi \left(2^2\right)=\mathrm{-8}\pi .\hfill \end{array}$

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