Answer to Question #125301 in Analytic Geometry for Samuel kassapa

Question

Find the area of the region inside both the rose r =sin(2Φ) and the circle r=cosΦ.The appropriate graph and intervals to the best of your interest to solve this question 

Answer 1

$$\sin(2\phi)=\cos(\phi)$$

$$2\sin(\phi)\cos(\phi)=\cos(\phi)$$

$$\cos(\phi)=0\text{ or } \sin(\phi)={1\over 2}$$

$$A_1={1\over2}\displaystyle\int_{0}^{\pi/2}(\sin(2\phi))^2d\phi=$$

$$={1\over4}\displaystyle\int_{0}^{\pi/2}((1-\cos(4\phi))d\phi=$$

$$={1\over4}\big[\phi-{1\over 4}\sin(4\phi)\big]\begin{matrix} \pi/2 \\ 0 \end{matrix}={\pi\over 8}(units^2)$$

$$A_2={1\over2}\displaystyle\int_{\pi/6}^{\pi/2}\bigg((\sin(2\phi))^2-(\cos\phi)^2\bigg)d\phi=$$

$$={1\over4}\displaystyle\int_{\pi/6}^{\pi/2}\bigg(1-\cos(4\phi)-1-\cos(2\phi)\bigg)d\phi=$$

$$=-{1\over4}\bigg[{1\over 4}\sin(4\phi)+{1\over 2}\sin(2\phi)\bigg]\begin{matrix} \pi/2 \\ \pi/6 \end{matrix}=$$

$$=-{1\over4}(0-{\sqrt{3}\over 8}+0-{\sqrt{3}\over 4})={3\sqrt{3}\over 32}(units^2)$$

$$Area=2(A_1-A_2)=2({\pi\over 8}-{3\sqrt{3}\over 32})={4\pi-3\sqrt{3}\over 16} (units^2)$$

$Area=\dfrac{4\pi-3\sqrt{3}}{16} \text{ square units}$