## Calculus: Early Transcendentals 8th Edition

(a) The length of the curve. $$y=2\ln (\sin \frac{1}{2}x), \quad\quad \frac{\pi}{3} \leq x \leq \pi.$$ is equal to $(\frac{21}{16} ).$ (b) The area of the surface obtained by rotating the curve in part (a) about the y-axis is equal to $(\frac{41}{10} \pi )$
(a) $$y=2\ln (\sin \frac{1}{2}x), \quad\quad \frac{\pi}{3} \leq x \leq \pi.$$ We have $$y=\frac{x^{4}}{16}+\frac{1}{2 x^{2}}=\frac{1}{16} x^{4}+\frac{1}{2} x^{-2} \Rightarrow \frac{d y}{d x}=\frac{1}{4} x^{3}-x^{-3}$$ $$\Rightarrow 1+(d y / d x)^{2}=1+\left(\frac{1}{4} x^{3}-x^{-3}\right)^{2}=1+\frac{1}{16} x^{6}-\frac{1}{2}+x^{-6}=\frac{1}{16} x^{6}+\frac{1}{2}+x^{-6}=\left(\frac{1}{4} x^{3}+x^{-3}\right)^{2}$$ Thus, the arc length is \begin{aligned} L& =\int_{1}^{2}\left(\frac{1}{4} x^{3}+x^{-3}\right) d x \\ & =\left[\frac{1}{16} x^{4}-\frac{1}{2} x^{-2}\right]_{1}^{2} \\ &=\left(1-\frac{1}{8}\right)-\left(\frac{1}{16}-\frac{1}{2}\right) \\ &=\frac{21}{16} \end{aligned} Hence, the exact length of the curve $$y=2\ln (\sin \frac{1}{2}x), \quad\quad \frac{\pi}{3} \leq x \leq \pi.$$ is equal to $(\frac{21}{16} ).$ (b) the area of the surface obtained by rotating the curve in part (a) about the y-axis is obtained by the following: \begin{aligned} S &=\int_{1}^{2} 2 \pi x\left(\frac{1}{4} x^{3}+x^{-3}\right) d x \\ &=2 \pi \int_{1}^{2}\left(\frac{1}{4} x^{4}+x^{-2}\right) d x \\ & =2 \pi\left[\frac{1}{20} x^{5}-\frac{1}{x}\right]_{1}^{2} \\ &=2 \pi\left[\left(\frac{32}{20}-\frac{1}{2}\right)-\left(\frac{1}{20}-1\right)\right] \\ &=2 \pi\left(\frac{8}{5}-\frac{1}{2}-\frac{1}{20}+1\right) \\ &=2 \pi\left(\frac{41}{20}\right) \\ &=\frac{41}{10} \pi \end{aligned} So, the area of the surface obtained by rotating the curve in part (a) about the y-axis is equal to $(\frac{41}{10} \pi )$