Answer
The area of the surface obtained by rotating the sine curve: $$ y= \sin x, \quad \quad 0 \leq x \leq \pi $$ about the x-axis is given by $$ \begin{aligned} S &=\int 2 \pi y d s\\ &\int_{0}^{\pi} 2 \pi ( \sin x) \sqrt{1+\left(\cos x \right)^{2}} dx \\ & \approx 14.426045 \quad\quad\left[\text {by using Simpson’s Rule } \right] \\ \end{aligned} $$
Work Step by Step
$$ y= \sin x, \quad \quad 0 \leq x \leq \pi $$ $\Rightarrow$ $$ y^{\prime} =\frac{dy}{dx}=\cos x $$ $\Rightarrow$ $$ \begin{aligned} ds &= \sqrt {1+\left(y^{\prime}\right)^{2}} dx \\ &=\sqrt{1+\left(\cos x \right)^{2}} dx\\ \end{aligned} $$ So, the area of the surface obtained by rotating the curve about the x-axis $$ \begin{aligned} S&=\int 2 \pi y d s \\ &=\int_{0}^{\pi} 2 \pi ( \sin x) \sqrt{1+\left(\cos x \right)^{2}} dx\\ \end{aligned} $$ To estimate the area of the surface obtained by rotating the sine curve by using Simpson’s Rule with the specified value of $ n=10$ , with $ n =10 , a = 0$, and $b = \pi $ we have $$ \Delta x=\frac{b-a}{n}=\frac{\pi-0}{10}=\frac{\pi}{10} $$ Let $$ f(x)=2 \pi ( \sin x) \sqrt{1+\left(\cos x \right)^{2}} $$ Then, $$ \begin{aligned} S &=\int_{0}^{\pi} 2 \pi ( \sin x) \sqrt{1+\left(\cos x \right)^{2}} dx\\ & = \frac{\Delta x}{3}\left[f\left(x_{0}\right)\right. +4 f\left(x_{1}\right)+2 f\left(x_{2}\right)+4 f\left(x_{3}\right)+\cdots \\ & \quad \left.+2 f\left(x_{n-2}\right)+4 f\left(x_{n-1}\right)+f\left(x_{n}\right)\right] \\ &\approx S_{10} \\ & =\frac{(\pi-0) / 10}{3}\left[g(0)+4 g\left(\frac{\pi}{10}\right)+2 g\left(\frac{2 \pi}{10}\right)+ 4 g\left(\frac{3 \pi}{10}\right)+2 g\left(\frac{4 \pi}{10}\right)\right.\\ &\left.+4 g\left(\frac{5 \pi}{10}\right)+2 g\left(\frac{6 \pi}{10}\right)+4 g\left(\frac{7 \pi}{10}\right)+2 g\left(\frac{8 \pi}{10}\right)+4 g\left(\frac{9 \pi}{10}\right)+g(\pi)\right] \\ & \approx 14.426045 \end{aligned} $$
