The problem asks us to find the volume generated by rotating the area bounded by the axes and the curve $y = cos(x)$ between $x=0$ to $x=\pi$ about the $x$-axis. We need to determine which of the provided options matches the calculated volume. There is an apparent typo in the prompt. Rather than through 2, it should read 'through $2\pi$'.

1. Problem Description

The problem asks us to find the volume generated by rotating the area bounded by the axes and the curve $y = cos(x)$ between $x=0$ to $x=\pi$ about the $x$-axis. We need to determine which of the provided options matches the calculated volume. There is an apparent typo in the prompt. Rather than through 2, it should read 'through $2\pi$'.

2. Solution Steps

We use the disk method to find the volume of the solid of revolution. The volume is given by the integral:

$V = \pi \int_{a}^{b} [f(x)]^2 dx$

In our case, $f(x) = cos(x)$, $a=0$, and $b=\pi$. Therefore, the volume is:

$V = \pi \int_{0}^{\pi} cos^2(x) dx$

We can rewrite $cos^2(x)$ using the identity:

$cos^2(x) = \frac{1 + cos(2x)}{2}$

So the integral becomes:

$V = \pi \int_{0}^{\pi} \frac{1 + cos(2x)}{2} dx$

$V = \frac{\pi}{2} \int_{0}^{\pi} (1 + cos(2x)) dx$

$V = \frac{\pi}{2} [x + \frac{1}{2}sin(2x)]_{0}^{\pi}$

$V = \frac{\pi}{2} [(\pi + \frac{1}{2}sin(2\pi)) - (0 + \frac{1}{2}sin(0))]$

$V = \frac{\pi}{2} [\pi + 0 - 0 - 0]$

$V = \frac{\pi}{2} * \pi$

$V = \frac{\pi^2}{2}$

None of the options matches $\frac{\pi^2}{2}$.

If instead, we consider the area bounded by the x-axis and the curve $y = cos(x)$ between $x=0$ and $x=\frac{\pi}{2}$, we would have:

$V = \pi \int_{0}^{\frac{\pi}{2}} cos^2(x) dx$

$V = \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}} (1 + cos(2x)) dx$

$V = \frac{\pi}{2} [x + \frac{1}{2}sin(2x)]_{0}^{\frac{\pi}{2}}$

$V = \frac{\pi}{2} [(\frac{\pi}{2} + \frac{1}{2}sin(\pi)) - (0 + \frac{1}{2}sin(0))]$

$V = \frac{\pi}{2} [\frac{\pi}{2} + 0 - 0 - 0]$

$V = \frac{\pi}{2} * \frac{\pi}{2}$

$V = \frac{\pi^2}{4}$

3. Final Answer

If the limits were 0 to $\pi/2$ then the answer is (D) $\frac{1}{4}\pi^2$