Answer
(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 $(\frac{41}{10} \pi )$
Work Step by Step
(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 $(\frac{41}{10} \pi )$
