Green's Theorem and the area between two circles Use Green's Theorem to evaluate the line integral:
\begin{align}\int_{C}(x-9y)dx+(x+y)dy\end{align}
C is the boundary of the region lying between the graphs: \begin{align}x^2+y^2=1\text{ and }x^2+y^2=81\end{align}
I understand that the easiest way would then be to find the area of each circle and subtract, giving a final answer of \begin{align}\ 800 \pi\end{align}
I would like to know how to work it out truly using Green's Theorem.  My thoughts are as follows:
\begin{align}\int_C \left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dA &=\int_{-9}^{9}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\left(1+9\right)dy\,dx \\\\&=20\int_{-9}^{9}\sqrt{1-x^2}dx\end{align}
This is where I am stuck.  I can use trig substitution but I end up with imaginary numbers.  I must be setting it up wrong but I am not sure where. I know my limits of integration are referencing the inner circle but the limits of integration on the exterior create the imaginary numbers.
Thank you for any assistance!
 A: In rectangular coordinates, the interior of $C$ (call it $D$) is the set
$$D=\left\{\begin{array}{c|c}
&\left(-9\le x\le-1\,\land\,|y|\le\sqrt{81-x^2}\right)\lor\\[1ex]
(x,y)&\left(|x|\le1\,\land\,\sqrt{1-x^2}\le|y|\le\sqrt{81-x^2}\right)\lor\\[1ex]&\left(1\le x\le9\,\land\,|y|\le\sqrt{81-x^2}\right)\end{array}\right\}$$
(there may be a way to simplify the inequalities but the above will do for your purposes)
The area of $D$ can be split up into $4$ integrals over the component regions $D_1,D_2,D_3,D_4$ so that
$$\iint_D\mathrm dA=\left\{\iint_{D_1}+\iint_{D_2}+\iint_{D_3}+\iint_{D_4}\right\}\mathrm dA$$

$$\begin{align*}
\iint_D\mathrm dA&=
\left\{\int_{-9}^{-1}\int_{-\sqrt{81-x^2}}^{\sqrt{81-x^2}}+\int_{-1}^1\int_{\sqrt{1-x^2}}^{\sqrt{81-x^2}}+\int_{-1}^1\int_{-\sqrt{81-x^2}}^{-\sqrt{1-x^2}}+\int_1^9\int_{-\sqrt{81-x^2}}^{\sqrt{81-x^2}}\right\}\mathrm dy\,\mathrm dx\\[1ex]
&=4\left\{\int_0^1\int_{\sqrt{1-x^2}}^{\sqrt{81-x^2}}+\int_1^9\int_0^{\sqrt{81-x^2}}\right\}\mathrm dy\,\mathrm dx\\[1ex]
&=4\int_0^1\sqrt{81-x^2}\,\mathrm dx+4\left\{\int_1^9-\int_0^1\right\}\sqrt{1-x^2}\,\mathrm dx
\end{align*}$$
where the second equality makes use of symmetry of these pieces of $D$.
Now you can try making those trig substitutions.

If you are at least a little familiar with polar coordinates, $D$ can be captured much more easily by the set
$$D=\left\{(r,\theta)\mid1\le r\le9\,\land\,0\le\theta\le2\pi\right\}$$
and the area of $D$ is trivial to compute:
$$\iint_D\mathrm dA=\int_0^{2\pi}\int_1^9r\,\mathrm dr\,\mathrm d\theta$$
