Answer to Question #195280 in Calculus for fraulyne rayne

Question

Find the common area enclosed by the following pairs of curves.

r^2=6cos2 theta and r=2cos theta

Answer 1

$\displaystyle r^2 = 6\cos^2(\theta),\,\, r=2\cos{\theta}\\ 6\cos^2(\theta) = 2\cos{\theta}\\ 2\cos{\theta}(3\cos{\theta} - 1) = 0\\ \cos{\theta} = 0,\,\ \theta = \frac{\pi}{2}\\ \cos{\theta} = \frac{1}{3},\,\, \theta = \arccos\left(\frac{1}{3}\right)\\ \begin{aligned} A_1 &= \frac{1}{2}\int_{\arccos\left(\frac{1}{3}\right)}^{\frac{\pi}{2}} 6\cos^2(\theta)\,\, \mathrm{d}\theta \\&= 3\int_{\arccos\left(\frac{1}{3}\right)}^{\frac{\pi}{2}} \cos^2(\theta)\,\, \mathrm{d}\theta \end{aligned}\\ \begin{aligned} A_2 &= \frac{1}{2} \int_{\arccos\left(\frac{1}{3}\right)}^{\frac{\pi}{2}} \left(2\cos{\theta}\right)^2 \,\, \mathrm{d}\theta \\&= 2\int_{\frac{\pi}{2}}^{\arccos\left(\frac{1}{3}\right)} \cos^2{\theta} \,\, \mathrm{d}\theta \end{aligned}\\ \begin{aligned} A &= A_1 - A_2 = 3\int_{\arccos\left(\frac{1}{3}\right)}^{\frac{\pi}{2}} \cos^2(\theta)\,\, \mathrm{d}\theta - 2\int_{\arccos\left(\frac{1}{3}\right)}^{\frac{\pi}{2}} cos^2{\theta} \,\, \mathrm{d}\theta \\&= \frac{\sin(2\theta) + 2\theta}{4}\biggr\vert_{\arccos\left(\frac{1}{3}\right)}^{\frac{\pi}{2}} \end{aligned}\\ \sin\left(\arccos{x}\right) = \sqrt{1 - x^2}\\ \sin\left(\arccos\left(\frac{1}{3}\right)\right) = \sqrt{1 - \frac{1}{9}} = \frac{2\sqrt{2}}{3}\\ \begin{aligned} \frac{\sin(2\theta) + 2\theta}{4}\biggr\vert_{\arccos\left(\frac{1}{3}\right)}^{\frac{\pi}{2}} &= \frac{-2\times \frac{2\sqrt{2}}{3} \times \frac{1}{3} - 2\arccos(1/3)}{4} + 2\frac{\pi/2}{4} \\&= \frac{-\frac{4\sqrt{2}}{9} + 2\textrm{csc}^{-1}(3)}{4} = \frac{\textrm{csc}^{-1}(3)}{2} - \frac{\sqrt{2}}{9} \end{aligned}\\ \therefore A = \frac{\textrm{csc}^{-1}(3)}{2} - \frac{\sqrt{2}}{9}.$