Answer
$$3.5846 $$
Work Step by Step
Given $$y=\cos x ; \quad\left[0, \frac{\pi}{2}\right] ; \quad y \text { -axis } ; \quad S_{8}$$
Since
$$ V= 2\pi \int_{0}^{\pi/2}x f(x)dx= 2\pi \int_{0}^{\pi/2} x\cos xdx$$
Now, we will evaluate the integral using $M_8$
Since $\Delta x=\dfrac{b-a}{n}=\dfrac{\pi/2}{8}=\dfrac{\pi}{16}$,
then by using Simpson’s rule, we get:
\begin{align*}
S_{n}&=\dfrac{\Delta x}{3}\left[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3).....+4f(x_{n-1})+f(x_n)\right]\\
S_{8}&=\dfrac{\Delta x}{3}\left[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+4f(x_5) +2f(x_6)+4f(x_7) +f(x_{8}) \right] \\
&=\dfrac{\pi}{48}\left[f(0)+4f(\pi/16)+2f(2\pi/16)+4f(3\pi/16)+2f(4\pi/16) +4f(5\pi/16)+2f(6\pi/16) +4f(7\pi/16)+f(\pi/2) \right]\\
&\approx 0.570799
\end{align*}
Hence $$ V=2\pi (0.570799)\approx 3.5846 $$
