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Green's theorem in the plane.

Theorem 8.20 (Flux-Divergence form)   Let $\overrightarrow{F} = M(x,y) \overrightarrow{i} + N(x,y) \overrightarrow{j} $ be a vector field over a region $\mathcal{D}$ in the plane. Denote by $\mathcal{C}$ a simple loop in $\mathcal{D}$, and by $\mathcal{R}$ the region enclosed by this loop.

The outward flux of $\overrightarrow{F} $ across $\mathcal{C}$ is equal to the double integral of $\text{div} \overrightarrow{F} $ over $\mathcal{R}$:

   \begin{displaymath}
\oint_{\mathcal{C}} \overrightarrow{F}\cdot \overrightarrow{...
...artial x} +\frac {\partial N}{\partial y} \right) \; dx \; dy.
\end{displaymath}

Example 8.21   We verify theorem  8.20 for $\overrightarrow{F} = (x+y) \overrightarrow{i} + (x-y) \overrightarrow{j} $ over the region $\mathcal{R}$ bounded by the circle whose parametrization is $\overrightarrow{r} (t)= \sin t \overrightarrow{i} + \cos t \overrightarrow{j} , \; 0 \leq t \leq 2 \pi$.

Theorem 8.22 (Circulation-Curl form)   Let $\overrightarrow{F} = M(x,y) \overrightarrow{i} + N(x,y) \overrightarrow{j} $ be a vector field over a region $\mathcal{D}$ in the plane. Denote by $\mathcal{C}$ a simple loop in $\mathcal{D}$, and by $\mathcal{R}$ the region enclosed by this loop.

The counterclockwise circulation of $\overrightarrow{F} $ along $\mathcal{C}$ is equal to the double integral of $\text{curl} \overrightarrow{F} $ over $\mathcal{R}$:


   \begin{displaymath}
\oint_{\mathcal{C}} \overrightarrow{F}\cdot \overrightarrow{...
...artial x} -\frac {\partial M}{\partial y} \right) \; dx \; dy.
\end{displaymath}

Example 8.23   We verify theorem  8.22 for $\overrightarrow{F} = (x+y) \overrightarrow{i} + (x-y) \overrightarrow{j} $ over the region $\mathcal{R}$ bounded by the circle whose parametrization is $\overrightarrow{r} (t)= \sin t \overrightarrow{i} + \cos t \overrightarrow{j} , \; 0 \leq t \leq 2 \pi$.

Example 8.24   Compute the integral $I=\int_\mathcal{C} (xy+y^2) \; dx + (x^2-y) dy$, where $\mathcal{C}$ is the border of the annulus $\mathcal{D}$ defined by $1 \leq x^2+y^2 \leq 9$.


  
Figure: The annulus: $1 \leq x^2+y^2 \leq 9$.
\begin{figure}
\mbox{\epsfig{file=Annulus.eps,height=5cm} }
\end{figure}

1.
Compute the line integral: The border $\mathcal{C}$ of the annulus is the union of two circles:
  • $\mathcal{C_1}$ whose equation is x2+y2=9; on $\mathcal{C_1}$, we turn counterclockwise, so a suitable parametrization is $x= 3 \cos t , \; y= 3 \sin t$, where $0 \leq t \leq 2 \pi.$
  • $\mathcal{C_2}$ whose equation is x2+y2=1; on $\mathcal{C_2}$, we turn clockwise, so a suitable parametrization is $x= \sin t , \; y= \cos t$, where $0 \leq t \leq 2 \pi.$
Now we compute the two line integrals:
\begin{align*}I_1 &=\int_\mathcal{C_1} (xy+y^2) \; ds \\
\quad &= \int_0^{2 \pi...
...+2 \right) \cos^2t - \cos t + \frac 23 \sin t \right]_0^{2 \pi }=0.
\end{align*}
Thus we have:

I=I1+I2= 0.

2.
With Green's theorem:

We have: M= xy+y2 and N=x2-y. Then:

\begin{displaymath}\frac {\partial N}{\partial x} = 2x \; \text{and} \; \frac {\partial M}{\partial y} =x+2y.
\end{displaymath}

We use theorem  8.22:
\begin{align*}I &= =\iint\limits_{\mathcal{R}} \left( \frac {\partial N}{\partia...
...t_0^{2 \pi} r^2 ( \cos \theta - 2 \sin \theta )
d \theta \; dr = 0.
\end{align*}


next up previous contents
Next: Surface integrals and Surface Up: Integration and Vector Fields. Previous: Flow, circulation and flux.
Noah Dana-Picard
2001-05-30