Examples

Area of a region calculated from the boundary

Area of a region calculated from the boundary

Set $\mathbf{F}=(0,x)$. Then $\frac{\partial F_2}{\partial x}=1$ while $\frac{\partial F_1}{\partial y}=0$, so the LHS of Green’s Theorem reduces to the enclosed area. We then have:

$$\text{Area}(\mathcal{D})={\oint }_{\mathcal{C}}xdy.$$

For example, an ellipse is given by $\mathbf{r}(t)=(a\cos (t),b\sin (t))$. Note $dy=b\cos (t)dt$. The area of an ellipse is therefore:

$$\text{Area}(\text{ellipse})={\int }_0^{2\pi }a\cos (t)b\cos (t)dt=\pi ab.$$

Faster calculation of line integral using Green’s Theorem

Faster calculation of line integral using Green’s Theorem

Problem: Find ${\oint }_{\mathcal{C}}{y}^{2}dx+3xydy$, where $\mathcal{C}$ is given as in the picture:

Solution: The interior of the region is covered by one double integral in polar coordinates, and that is easier to compute than four line integrals, so we use Green’s Theorem to convert to a double integral. We choose the vector field $\mathbf{F}=(y^2,3xy)$ so that $\mathbf{F}\cdot d\mathbf{s}$ gives the line element. Then $\nabla {\times }_z\mathbf{F}=\frac{\partial }{\partial x}(3xy)-\frac{\partial }{\partial y}{y}^2=y$, and we compute:

$$\begin{array}{rl}{\oint }_{\mathcal{C}}{y}^{2}dx+3xydy& ={\iint }_{\mathcal{D}}\nabla {\times }_z\mathbf{F}d\mathbf{s}={\iint }_{\mathcal{D}}ydA\\ & ={\int }_0^{\pi }{\int }_1^{2}r\sin (\theta )rdrd\theta =\frac{14}{3}.\end{array}$$