Hi. I wanted to know if I did this right:

Using double integrals, set in polar coordinates, the integral that gives the area for the region $\displaystyle D: \{x^2+y^2\leq{9},(x-2)^2+y^2\geq{7}$

From the picture:

The pink region represents the area I want.

The intersection is given for x=3/2, where: $\displaystyle y=\displaystyle\frac{3\sqrt[ ]{3}}{2}$

Then $\displaystyle \tan \theta_1=9\sqrt[ ]{3}$

Tita 1 represents the limit on the angle for the circle that isn't centered at the origin. Is the limit for the brown region.

$\displaystyle \tan \theta_2=3\displaystyle\frac{\sqrt[ ]{3}}{4}$

This is the limit on the angle for the orange region, for the circle centered at the origin.

Finally

$\displaystyle y=\displaystyle\frac{-3\sqrt[ ]{3}}{2}(x+3)$

With $\displaystyle x=rcos\theta,y=r\sin \theta$

Then $\displaystyle r=\frac{-15\sqrt[ ]{3}}{2(\sin\theta+\frac{3\sqrt[ ]{3}}{2}\cos\theta)}$

And the area:

$\displaystyle \displaystyle\int_{0}^{2\pi}\int_{0}^{3}rdrd\theta-2\displaystyle\int_{0}^{tan^{-1}9\sqrt[ ]{3}}\int_{0}^{\sqrt[ ]{7}}rdrd\theta-2\displaystyle\int_{0}^{tan^{-1}\displaystyle\frac{3\sqrt[ ]{3}}{4}}\int_{\frac{-15\sqrt[ ]{3}}{2(\sin\theta+\frac{3\sqrt[ ]{3}}{2}\cos\theta)}}^{3}rdrd\theta$

The first integral represents the entire circle, the second the brown region and the third the orange region, both multiplied on two by simmetry