# Thread: Area of region bounded by polar equation

1. ## Area of region bounded by polar equation

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)

2. Originally Posted by clev88
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 ${r_1} = 2,{\text{ }}{r_2} = 4\cos \theta ,{\text{ }}{S_{{r_2}}}\backslash \left\{ {{S_{{r_1}}} \cap {S_{{r_2}}}} \right\} = ?$

Let $S = {S_{{r_2}}}\backslash \left\{ {{S_{{r_1}}} \cap {S_{{r_2}}}} \right\}$.

Find the intersection points of polar curves

${r_1} = {r_2} \Leftrightarrow 2 = 4\cos \theta \Leftrightarrow \cos \theta = \frac{1}{2} \Leftrightarrow \theta = \pm \frac{\pi }{3}{\text{ }}\left( {k = 0} \right).$

Then, taking into account the symmetry, we have

$S = 2\left[ {\frac{1}{2}\int\limits_0^{\pi /3} {r_2^2d\theta } - \frac{1}{2}\int\limits_0^{\pi /3} {r_1^2d\theta } } \right] = \int\limits_0^{\pi /3} {\left( {r_2^2 - r_1^2} \right)d\theta } = \int\limits_0^{\pi /3} {\left( {{{\left( {4\cos \theta } \right)}^2} - {2^2}} \right)d\theta } =$

$= \int\limits_0^{\pi /3} {\left( {16{{\cos }^2}\theta - 4} \right)d\theta } = \int\limits_0^{\pi /3} {\left( {8\left( {1 + \cos 2\theta } \right) - 4} \right)d\theta } = 4\int\limits_0^{\pi /3} {\left( {1 + 2\cos 2\theta } \right)d\theta } =$

$= 4\left. {\left( {\theta + \sin 2\theta } \right)} \right|_0^{\pi /3} = 4\left( {\frac{\pi }{3} + \frac{{\sqrt 3 }}{2}} \right) = \frac{{4\pi }}{3} + 2\sqrt 3 .$

See picture

3. ## thanks

Thank you this helps!