Question

Find the area of the region that lies inside the first curve and outside the second curve.
r = 11cos θ, r = 5 + cos θ

Answer 1

i got 20sqrt(3)+35/3 pi, but its apparently wrong
i no its close though

Answer 2

http://en.wikipedia.org/wiki/Polar_equation#Integral_calculus_.28area.29
using this general integral apply the fact that we want area of region curve1 - curve2
the curves intersect at theta = +-pi/3
\[A = 2\int\limits_{0}^{\pi/3}\frac{1}{2}(11\cos \theta)^{2} - (5+\cos \theta)^{2} d \theta\]
\[=\int\limits_{0}^{\pi/3}120\cos^{2} \theta-10\cos \theta -25\]
\[=10\sqrt{3}+\frac{35\pi}{3}\]