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• Notes

Draw the curve $r=1+2 \cos \theta$

Find the area that lies inside the curves $r=\cos \theta$ and $r=1-\cos \theta$

$r=r \Rightarrow \cos \theta=1-\cos \theta \Rightarrow \cos \theta+\cos \theta=1$

$\Rightarrow 2 \cos \theta=1 \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=\cos ^{1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$

Area $=\int_{\alpha}^{\beta} \frac{1}{2} r^{2} d \theta$

$A_{1}= \int_{0}^{\pi / 3}\frac{1}{2}(1-\cos \theta)^{2} d \theta=\frac{1}{2} \int_{0}^{\pi / 3}\left(1-2 \cos \theta+\cos ^{2} \theta\right) d \theta$

$\cos 2 \theta=2 \cos ^{2} \theta-1$

$2 \cos ^{2} \theta=\cos 2 \theta+1$

$\frac {2 \cos ^{2} \theta}{2}=\frac{\cos 2\theta+1}{2}$

$\cos ^{2} \theta=\frac{1}{2} \cos 2 \theta+\frac{1}{2}$

$A_{1}=\frac{1}{2} \int_{0}^{\frac{\pi}{3}}\left(1-2 \cos \theta+\frac{1}{2}+\frac{1}{2} \cos 2 \theta\right) d \theta$

$A_{1}=\frac{1}{2} \int_{0}^{\pi / 3}\left(\frac{3}{2}-2 \cos \theta+\frac{1}{2} \cos 2 \theta\right) d \theta$

$A_{1}=\frac{1}{2}\left[\frac{3}{2} \theta-2 \sin \theta+\frac{1}{2} \cdot \frac{1}{2} \sin 2 \theta\right]_{0}^\frac {\pi}{3}$

$A_{1}=\frac{1}{2}\left[\frac{3}{2}\left(\frac{\pi}{3}\right)-2 \sin \frac{\pi}{3}+\frac{1}{4} \sin 2\left(\frac{\pi}{3}\right)-(0)\right]$

$A_{1}=\frac{1}{2}\left[\frac{\pi}{2}-2\left(\frac{\sqrt{3}}{2}\right)+\frac{1}{4} \cdot\left(\frac{\sqrt{3}}{2}\right)\right]=\frac{\pi}{4}-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{16}$

$A_{1}=\frac{\pi}{4}-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{16}$

$A_{2}=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{1}{2}(\cos \theta)^{2} d \theta$

$=\frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \cos ^{2} \theta d \theta=\frac{1}{2} \int_{\frac{\pi}{3}}^\frac {\pi}{2}\left(\frac{1}{2} \cos 2 \theta+\frac{1}{2}\right) d \theta$

$A_{2}=\frac{1}{4} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}(\cos 2 \theta+1) d \theta=\frac{1}{4} \cdot\left[\frac{1}{2} \sin 2 \theta+\theta\right]_{\pi / 3}^{\pi / 2}$

$A_{2}=\frac{1}{4}\left[\frac{1}{2} \sin \left(2 \cdot \frac{\pi}{2}\right)+\frac{\pi}{2}-\left(\left(\frac{1}{2} \sin 2\left(\frac{\pi}{3}\right)+\frac{\pi}{3}\right)\right]\right.$

$A_{2}=\frac{1}{4}\left[0+\frac{\pi}{2}-\left(\frac{\sqrt{3}}{4}+\frac{\pi}{3}\right)\right]=\frac{\pi}{8}-\frac{\sqrt{3}}{16}-\frac{\pi}{12}$

$A_{1}=\frac{\pi}{4}-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{16}$

$A_{2}=\frac{\pi}{8}-\frac{\sqrt{3}}{16}-\frac{\pi}{12}$

Area $=2\left(A_{1}+A_{2}\right)=2\left(\frac{\pi}{4}-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{16}+\frac{\pi}{8}-\frac{\sqrt{3}}{16}-\frac{\pi}{12}\right)$

$=\frac{\pi}{2}-\sqrt{3}+\frac{\pi}{4}-\frac{\pi}{6}$

$=\frac{6 \pi}{12}-\frac{12 \sqrt{3}}{12}+\frac{3 \pi}{12}-\frac{2 \pi}{12}$

Area $=\frac{7 \pi-12 \sqrt{3}}{12}$