6.4 Green’s theorem  (Page 4/12)

Let D be the disk enclosed by C. The flux across C is ${\displaystyle {\oint }_C\text{F}·\text{N}ds}.$ We could evaluate this integral using tools we have learned, but Green’s theorem makes the calculation much more simple. Let $P\left(x,y\right)=x$ and $Q\left(x,y\right)=y$ so that $\text{F}=⟨P,Q⟩.$ Note that $P_x=1=Q_y,$ and therefore $P_x+Q_y=2.$ By Green’s theorem,

Since ${\displaystyle \int {\displaystyle {\int }_{D}dA}}$ is the area of the circle, ${\displaystyle \int {\displaystyle {\int }_{D}dA}}=\pi r^2.$ Therefore, the flux across C is $2\pi r^2.$

To calculate the flux without Green’s theorem, we would need to break the flux integral into three line integrals, one integral for each side of the triangle. Using Green’s theorem to translate the flux line integral into a single double integral is much more simple.

Let D be the region enclosed by S . Note that $P_x=2x$ and $Q_y=1;$ therefore, $P_x+Q_y=2x+1.$ Green’s theorem applies only to simple closed curves oriented counterclockwise, but we can still apply the theorem because ${\displaystyle {\oint }_C\text{F}·\text{N}ds=\text{−}}{\displaystyle {\oint }_{\text{−}S}\text{F}·\text{N}ds}$ and $\text{−}S$ is oriented counterclockwise. By Green’s theorem, the flux is

Notice that the top edge of the triangle is the line $y=\mathrm{-3}x+3.$ Therefore, in the iterated double integral, the y -values run from $y=0$ to $y=\mathrm{-3}x+3,$ and we have

Let C represent the given rectangle and let D be the rectangular region enclosed by C . To find the amount of water flowing across C , we calculate flux ${\displaystyle {\int }_C\text{v}•d\text{r}.}$ Let $P\left(x,y\right)=5x+y$ and $Q\left(x,y\right)=x+3y$ so that $\text{v}=\left(P,Q\right).$ Then, $P_x=5$ and $Q_y=3.$ By Green’s theorem,

Therefore, the water flux is 80 m ² /sec.