# Area of region bounded by polar equation

• July 29th 2009, 09:23 PM
clev88
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)
• July 30th 2009, 12:34 AM
DeMath
Quote:

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