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!!
