# Green's Theorem and limits on y for flux

I'm working through understanding the example provided in the book for the divergence integral. The theorem (Green's):

$$\oint_C = \mathbf{F}\cdot \mathbf{T}ds = \oint_CMdy-Ndx=\int\int_R(\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y} )dxdy$$

The example uses the following: $\mathbf{F}(x,y) = (x-y)\mathbf{i} + x\mathbf{j}$ over the region $\mathbf{R}$ bounded by the unit circle $C: \mathbf{r}(t)=cos(t)\mathbf{i} + sin(t)\mathbf{j}, 0 \le t \le 2\pi$.

There is then the following relations:

$$\begin{array}{rr} M = cos(t) - sin(t) & dx = d(cos(t)) = -sin(t)dt \\ N = cos(t) & dy = d(sin(t)) = cos(t)dt \end{array} \\ \begin{array}{rrrr} \frac{\partial M}{\partial x}=1 & \frac{\partial M}{\partial y} = -1 & \frac{\partial N}{\partial x}=1 & \frac{\partial N}{\partial y} = 0 \end{array}$$

Now that the foundation is laid, here's the rightmost part of the first equation given:

$$\begin{array}{rcl} \int\int_R \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} dxdy & = & \int\int_R 1 + 0 dxdy \\ & = & \int\int_R dxdy \\ & = & \text{area inside unit circle} \\ & = & \pi \end{array}$$

I understand it intuitively because it's the area over that region and the area of a circle is $A = \pi\cdot r^2$. With $r = 1$ that's obviously $\pi$. What I'm not sure of is how to express it in an integral. The question $x^2 + y^2 = 1$ represents the unit circle. Thus, $x$ as a function of $y$ I get $x = \sqrt{1-y^2}$, thus the final stage I show should be:

$$\int_{?}^{?} \int_{0}^{\sqrt{1-y^2}}dx dy$$

right? What should be used for the limits on y? I know it's simple but I'm just not seeing it and I need some guidance.

Thanks

For Green's Theorem

$$\oint_C (Mdx+Ndy)=\int \int_R \left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right) dxdy$$

Here, $M=(x-y)$ and $N=x$ such that

\begin{align}\oint_C (Mdx+Ndy)&=2\int \int_R dxdy\\\\ &=2\int_{-1}^{1} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}dxdy\\\\ &=4\int_{-1}^{1} \sqrt{1-y^2}dy\\\\ &=2\left(y\sqrt{1-y^2}+\arcsin(y)\right)|_{-1}^{1}\\\\ &=2(\arcsin(1)-\arcsin(-1))\\\\ &=2\pi \end{align}

If we evaluate the line integral in a straightforward way, we let $x=\cos \phi$ and $y=\sin \phi$. Then, $dx=-\sin \phi d\phi$ and $dy=\cos \phi d\phi$. We obtain the following

\begin{align}\oint_C (Mdx+Ndy)&=\int_0^{2\pi} (-\cos \phi \sin \phi+\sin^2 \phi+\cos^2 \phi)d\phi\\\\ &=\int_0^{2\pi} (-\cos \phi \sin \phi+1)d\phi\\\\ &=2\pi \end{align}

as expected!!

• Thanks. This makes sense. I knew it had to be something simple. – Andrew Falanga Apr 19 '15 at 16:56