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!
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:
$$\iint_D\mathrm dA=\int_0^{2\pi}\int_1^9r\,\mathrm dr\,\mathrm d\theta$$