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Find the area enclosed by the curve \(r=\cos (2 \theta)\)

\(A_{1}=\int_{0}^{\pi / 4} \frac{1}{2} r^{2} d \theta=\int_{0}^{\pi / 4} \frac{1}{2}(\cos (2\theta))^{2} d \theta\)

\(A_{1}=\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \cos ^{2}(2 \theta) d \theta\)

\(\cos ^{2} \theta=\frac{1}{2}+\frac{1}{2} \cos (2\theta)\)

\(\cos ^{2} 2 \theta=\frac{1}{2}+\frac{1}{2} \cos (4 \theta)\)

\(A_{1}=\frac{1}{2} \int_{0}^{\frac {\pi} {4}}\left(\frac{1}{2}+\frac{1}{2} \cos (4 \theta)\right) d \theta\)

\(A_{1}=\frac{1}{4} \int_{0}^{\pi / 4}(1+\cos 4 \theta) d \theta=\frac{1}{4}\left[\theta+\frac{1}{4} \sin (4\theta)\right]_{0}^{\pi / 4}\)

\(A_{1}=\frac{1}{4}\left[\frac{\pi}{4}+\frac{1}{4} \sin \left(4 \cdot \frac{\pi}{4}\right)-(0)\right]=\frac{\pi}{16}\)

Area \(=8 A_{1}=8 \cdot \frac{\pi}{16}=\frac{\pi}{2}\)

Find the area inside \(r=2+2 \cos \theta\) and outside \(r=3\)

\(r=r \rightarrow 2+2 \cos \theta=3 \rightarrow 2 \cos \theta=3-2 \rightarrow 2 \cos \theta=1\)

\(\rightarrow \cos \theta=\frac{1}{2} \rightarrow \theta=\cos^{-1} \left(\frac{1}{2}\right)=\pm \frac{\pi}{3}\)

\(A=\frac{1}{2} \int_{-\frac{\pi}{3}}^{\pi / 3}(2+2 \cos \theta)^{2}-(3)^{2} ) d \theta\)

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

\(A=2 \cdot \frac{1}{2} \int_{0}^{\frac {\pi} {3}}\left(4+8 \cos \theta+4 \cos ^{2} \theta-9\right) d \theta\)

\(A=\int_{0}^{\pi / 3}\left(8 \cos \theta+4 \cos ^{2} \theta-5\right) d \theta\)

\(\cos ^{2} \theta=\frac{1}{2}+\frac{1}{2} \cos 2 \theta\)

\(A=\int_{0}^{\frac{\pi}{3}}\left(8 \cos \theta+4 \cdot\left(\frac{1}{2}+\frac{1}{2} \cos 2 \theta\right)-5\right) d \theta\)

\(A=\left[8 \sin \theta+2 \theta+2 \frac{1}{2} \sin 2 \theta-5 \theta\right]_{0}^\frac {\pi}{3}\)

\(A=[8 \sin \theta-3 \theta+\sin 2 \theta]_{0}^{\frac{\pi}{3}}\)

\(A=\left[\left(8 \sin \frac{\pi}{3}-3 \cdot \frac{\pi}{3}+\sin 2 \frac{\pi}{3}\right)-(0)\right]\)

\(A=8 \frac{\sqrt{3}}{2}-\pi+\frac{\sqrt{3}}{2}\)

\(=4 \sqrt{3}-\pi+\frac{\sqrt{3}}{2}\)

\(=\frac{8 \sqrt{3}}{2}+\frac{\sqrt{3}}{2}-\pi=\frac{9 \sqrt{3}}{2}-\pi\)

Find the length of the polar curve \(r=e^{2 \theta}\) where \(0 \leq \theta \leq 2 \pi\)

arc length \(=\int_{0}^{2 \pi} \sqrt{r^{2}+\left(\frac{d r}{d \theta}\right)^{2} d \theta}\)

\(r=e^{2 \theta} \rightarrow \frac{d r}{d \theta}=2 \cdot e^{2 \theta}\)

\(\rightarrow\left(\frac{d r}{d \theta}\right)^{2}=\left(2 e^{2 \theta}\right)^{2}=4 e^{4 \theta}\)

arc length \(=\int_{0}^{2 \pi} \sqrt{\left(e^{2\theta}\right)^{2}+4 e^{4\theta}} d \theta\)

\(=\int_{0}^{2 \pi} \sqrt{e^{4\theta}+4 e^{4\theta}} d \theta\)

arc length \(=\int_{0}^{2 \pi} \sqrt{5 e^{4 \theta}} d \theta=\int_{0}^{2 \pi} e^{2 \theta} \cdot \sqrt{5} d \theta\)

\(=\sqrt{5} \int_{0}^{2 \pi} e^{2\theta} d \theta\)

arc length \(=\sqrt{5} \cdot e^{2\theta} \cdot\left.\frac{1}{2}\right|_{0}^{2 \pi}\)

\(=\frac{\sqrt{5}}{2} e^{2 \cdot 2 \pi}-\frac{\sqrt{5}}{2} e^{2 \cdot(0)}=\frac{\sqrt{5}}{2}  [e^{4 \pi} - 1]\)

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