Answer
$V\approx 2.4649$
Work Step by Step
Given $$y=\cos x ; \quad\left[0, \frac{\pi}{2}\right] ; \quad x \text { -axis; } \quad M_{8}$$
Since
$$ V= \pi \int_{0}^{\pi/2} [f(x)]^2dx= \pi \int_{0}^{\pi/2} \cos^2xdx$$
Now, we will evaluate the integral using $M_8$, since $\Delta x=\dfrac{b-a}{n}=\dfrac{\pi/2}{8}=\dfrac{\pi}{16}$
Therefore, the sub intervals consist of
$$[0,\pi/16],[\pi/16, 2\pi/16],[2\pi/16],\cdots ,[7\pi/16,\pi/2] $$
The midpoints of these sub intervals are
$$\{ \pi/32,\ 3\pi/32,\ \ \cdots 15\pi/32 \}$$
Hence
\begin{align*}
M_{n}&= \sum_{i=1}^{n}f(m_i)\Delta x\\
M_{4}&= \left[ f(m_1)+ f(m_2)+ ..............+f(m_{8})\right]\Delta x\\
&=\frac{\pi}{32}\left[ f(\pi/32)+ f(3\pi/32)+\cdots+ f(15\pi/32)\right]\\
&\approx 0.785
\end{align*}
Hence, $V\approx 2.4649$
