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Find the arc length of the polar curve $r=\theta^{2}$ where $0 \leq \theta \leq \pi$

$r=\theta^{2}$

$\frac{d r}{d \theta}=2\theta$

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

$=\int_{0}^{\pi} \sqrt{\left(\theta^{2}\right)^{2}+(2\theta)^{2}} d \theta$

$L=\int_{0}^{\pi} \sqrt{\theta^{4}+4 \theta^{2}} d \theta=\int_{0}^{\pi} \sqrt{\theta^{2}\left(\theta^{2}+4\right)} d \theta$

$=\int_{0}^{\pi} \theta \sqrt{\theta^{2}+4} d \theta$

let $\theta^{2}+4=t \rightarrow d t=2 \theta d \theta$

$\theta=0 \rightarrow t=4$

$\theta=\pi \rightarrow t=\pi^{2}+4$

$L=\frac{1}{2} \int_{4}^{\pi^{2}+4} \sqrt{t} \cdot d t=\frac{1}{2} \int_{4}^{\pi^{2}+4} t^{\frac{1}{2}} d t$

$=\frac{1}{2} \cdot\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right]_{4}^{\pi^{2}+4}=\frac{1}{2} \times \frac{2}{3}\left[t^{\frac{3}{2}}\right]_{4}^{\pi^{2}+ 4}$

$=\frac{1}{3}\left[\sqrt{\left(\pi^{2}+4)^{3}\right.}-\sqrt{(4)^{3}}\right]=\frac{1}{3}\left[\sqrt{\left(\pi^{2}+4\right)^{2}}-8\right]$

Find the arc length of the polar curve $r=\cos ^{2}\left(\frac{\theta}{2}\right), \theta \in[0,2 \pi]$

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

$\frac{d r}{d \theta}=\frac{d\left(\cos ^{2} \frac{\theta}{2}\right)}{d \theta}$

$=2 \cos \frac{\theta}{2} \cdot\left(-\sin \frac{\theta}{2}\right)\left(\frac{1}{2}\right)=-\cos \left(\frac{\theta}{2}\right) \cdot \sin \left(\frac{\theta}{2}\right)$

$\left(\frac{d r}{d \theta}\right)^{2}=$

$=\left(2 \cos \frac{\theta}{2} \cdot\left(-\sin \frac{\theta}{2}\right)\left(\frac{1}{2}\right)\right)^{2}=\left(-\cos \left(\frac{\theta}{2}\right) \cdot \sin \left(\frac{\theta}{2}\right)\right)^{2}$

$=\cos ^{2} \frac{\theta}{2} \sin \frac{2\theta}{2}$

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

$=\int_{0}^{2 \pi} \sqrt{\cos ^{2} \frac{\theta}{2}\left(\cos ^{2} \frac{\theta}{2}+\sin \frac{2 \theta}{2}\right)} d \theta$

arc length $=\int_{0}^{2 \pi} \sqrt{\cos ^{2} \frac{\theta}{2}} d \theta=\int_{0}^{2 \pi}\left|\cos \left(\frac{\theta}{2}\right)\right| d \theta$

$\rightarrow+\quad \theta \in[0, \pi]$

$\rightarrow-\theta \in[\pi , 2 \pi]$

arc length $=\int_{0}^{2 \pi}\left|\cos \frac{\pi}{2}\right| d \theta=\int_{0}^{\pi} \cos \frac{\theta}{2} d \theta+\int_{\pi}^{2 \pi}-\cos \frac{ \theta}{2} d \theta$

$=2 \cdot \sin \left.\frac{\theta}{2}\right|_{0}^{\pi}-\left.\left(2 \cdot \sin \frac{\theta}{2}\right)\right|_{\pi} ^{2 \pi}$

$=2(1)-(0)-\left(2 (0)-2 \cdot(1)\right)$

$=2+2=4$