The problem asks us to find the volume generated by rotating the area bounded by the axes and the curve $y = cos(x)$ between $x=0$ and $x=\pi$ about the x-axis.

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 $x=0$ and $x=\pi$ about the x-axis.

2. Solution Steps

We will use the disk method to find the volume of the solid generated by rotating the area under the curve $y=cos(x)$ about the x-axis, from $x=0$ to $x=\pi$. The volume element is given by the area of the disk, $\pi y^2$, times the thickness $dx$. Thus, we integrate $\pi y^2$ with respect to $x$ from $0$ to $\pi$:

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

We can use the identity $cos^2(x) = \frac{1 + cos(2x)}{2}$ to simplify the integral:

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

Now, we integrate term by term:

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

Since $sin(2\pi) = 0$ and $sin(0) = 0$:

$V = \frac{\pi}{2} (\pi + 0 - 0 - 0) = \frac{\pi}{2} (\pi) = \frac{\pi^2}{2}$

3. Final Answer

The volume is $\frac{\pi^2}{2}$.