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Find the arc length of the curve defined by the parametric equations:

$x=t-\sin t, y=1-\cos t, 0 \leq t \leq 2 \pi$

arc length $=\int_{0}^{2 \pi} \sqrt{\left(\frac{d x}{dt}\right)^{2}+\left(\frac{d y}{dt}\right)^{2}} d t$

$\frac{d x}{d t}=1-\cos t \Rightarrow\left(\frac{d x}{d t}\right)^{2}=(1-\cos t)^{2}=1-2 \cos t+\cos ^{2} t$

$\frac{d y}{d t}=\sin t \quad \Rightarrow\left(\frac{d y}{d t}\right)^{2}=(\sin (t))^{2}=\sin ^{2}(t)$

Arc length $=\int_{0}^{2 \pi} \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t=\int_{0}^{2 \pi} \sqrt{1-2 \cos t+\cos ^{2} t+\sin ^{2}(t)} d t$

Arc length $=\int_{0}^{2 \pi} \sqrt{2-2 \cos t} d t=\int_{0}^{2 \pi} \sqrt{2(1-\cos t)} d t=\sqrt{2} \int_{0}^{2 \pi} \sqrt{1-\cos t} d t$

$\sin ^{2}(t / 2)=\frac{1}{2}(1-\cos t) \rightarrow 1-\cos t=2 \sin ^{2}(t / 2)$

$=\sqrt{2} \int_{0}^{2 \pi} \sqrt{2 \sin ^{2}(t / 1)} d t=\sqrt{2} \cdot \sqrt{2} \int_{0}^{2 \pi} \sqrt{\sin ^{2}(t / 2)} d t$

$=2 \int_{0}^{2 \pi} \sin (t / 2) d t=2 \cdot[-\cos (t / 2) \cdot 2]_{0}^{2 \pi}$

$=-4[\cos (t / 2)]_{0}^{2 \pi}=-4\left[\cos \left(\frac{2 \pi}{2}\right)-\cos (0)\right]=-4[-1-1]=8$

Find the area of surface of revolution obtained by rotating the curve defined by the parametric equations:

$x=\cos ^{3} t, y=\sin ^{3} t, 0 \leq t \leq \frac{\pi}{2}$

surface area $=s=\int_{0}^{\pi / 2} 2 \pi y \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t$

$\left(\frac{d x}{d t}\right)^{2}=\left(-3 \cos ^{2}+\sin t\right)^{2}=9 \cos ^{4} t \sin ^{2} t$

$\left(\frac{d y}{d t}\right)^{2}=\left(3 \sin ^{2}+\cos t\right)^{2}=9 \sin ^{4} t \cos ^{2} t$

$\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}=9 \cos ^{4} t \sin ^{2} t+9 \sin ^{4} t \cos ^{2} t=9 \sin ^{2} t \cos ^{2}+\left(\cos ^{2} t+4 \sin ^{2} t\right)$

$=S=\int_{0}^{\pi / 2} 2 \pi \cdot \sin ^{3} t \sqrt{9 \sin ^{2}t \cos ^{2} t} dt=\int_{0}^{\pi / 2} 6 \pi \sin ^{3} t \sqrt{\sin ^{2} t \cos ^{2} t} dt$

$=\int_{0}^{2 \pi} 6\pi \sin ^{3} t \cdot \sin t \cos t dt$

$S=6 \pi \int_{0}^{\pi / 2} \sin ^{4} t \cos t d t$

$\Rightarrow let \: u=\sin t \Rightarrow d u=\cos t d t$

when $t=0 \Rightarrow u=\sin (0)=0$

when $t=\frac{\pi}{2} \Rightarrow u=\sin \left(\frac{\pi}{2}\right)=1$

$s=6 \pi \int_{0}^{1} u^{4} d u=6 \pi\left[\frac{u^{5}}{5}\right]_{0}^{1}$

$=6 \pi\left[\frac{(1)^5}{5}-(0)\right]=6 \pi \cdot \frac{1}{5}=\frac{6 \pi}{5}$