Find the area of the region that is outside the graph of the first equation and inside the graph of the second equation.
1. r=2
2. r= 4cosx
(x = theta)
So $\displaystyle {r_1} = 2,{\text{ }}{r_2} = 4\cos \theta ,{\text{ }}{S_{{r_2}}}\backslash \left\{ {{S_{{r_1}}} \cap {S_{{r_2}}}} \right\} = ?$
Let $\displaystyle S = {S_{{r_2}}}\backslash \left\{ {{S_{{r_1}}} \cap {S_{{r_2}}}} \right\}$.
Find the intersection points of polar curves
$\displaystyle {r_1} = {r_2} \Leftrightarrow 2 = 4\cos \theta \Leftrightarrow \cos \theta = \frac{1}{2} \Leftrightarrow \theta = \pm \frac{\pi }{3}{\text{ }}\left( {k = 0} \right).$
Then, taking into account the symmetry, we have
$\displaystyle S = 2\left[ {\frac{1}{2}\int\limits_0^{\pi /3} {r_2^2d\theta } - \frac{1}{2}\int\limits_0^{\pi /3} {r_1^2d\theta } } \right] = \int\limits_0^{\pi /3} {\left( {r_2^2 - r_1^2} \right)d\theta } = \int\limits_0^{\pi /3} {\left( {{{\left( {4\cos \theta } \right)}^2} - {2^2}} \right)d\theta } =$
$\displaystyle = \int\limits_0^{\pi /3} {\left( {16{{\cos }^2}\theta - 4} \right)d\theta } = \int\limits_0^{\pi /3} {\left( {8\left( {1 + \cos 2\theta } \right) - 4} \right)d\theta } = 4\int\limits_0^{\pi /3} {\left( {1 + 2\cos 2\theta } \right)d\theta } =$
$\displaystyle = 4\left. {\left( {\theta + \sin 2\theta } \right)} \right|_0^{\pi /3} = 4\left( {\frac{\pi }{3} + \frac{{\sqrt 3 }}{2}} \right) = \frac{{4\pi }}{3} + 2\sqrt 3 .$
See picture