To find the volume of the solid generated by rotating the region bounded by $y = \sin(3x)$, $y = 0$, $x = 0$, and $x = \frac{\pi}{3}$ about the $x$-axis, we use the method of disks. The formula for the volume $V$ when the region between the curve $y = f(x)$ and the $x$-axis from $x = a$ to $x = b$ is rotated about the $x$-axis is given by:

$V = \pi \int_a^b [f(x)]^2 \, dx$

For the given function $f(x) = \sin(3x)$, the limits of integration are $a = 0$ and $b = \frac{\pi}{3}$. Substituting these into the volume formula, we have:

$V = \pi \int_0^{\frac{\pi}{3}} [\sin(3x)]^2 \, dx$

To simplify the integral, we can use the trigonometric identity $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$:

$[\sin(3x)]^2 = \frac{1 - \cos(6x)}{2}$

The volume integral becomes:

$V = \pi \int_0^{\frac{\pi}{3}} \frac{1 - \cos(6x)}{2} \, dx$

This simplifies to:

$V = \frac{\pi}{2} \int_0^{\frac{\pi}{3}} (1 - \cos(6x)) \, dx$

We can now compute this integral:

$\int_0^{\frac{\pi}{3}} 1 \, dx = \frac{\pi}{3}$

$\int_0^{\frac{\pi}{3}} \cos(6x) \, dx = \frac{1}{6} \sin(6x) \Big|_0^{\frac{\pi}{3}} = \frac{1}{6}(\sin(2\pi) - \sin(0)) = 0$

Thus, the volume is:

$V = \frac{\pi}{2} \left(\frac{\pi}{3} - 0\right) = \frac{\pi^2}{6}$

The volume of the solid generated is $\frac{\pi^2}{6}$ cubic units.

Would you like more details on this calculation or help with another problem?

Here are some related questions to consider:

1. What would be the volume if the same region were rotated about the y-axis?
2. How does changing the limits of integration affect the volume of the solid?
3. What would the volume be if $y = \sin(4x)$ instead?
4. Can you calculate the volume using cylindrical shells instead of disks for this problem?
5. How does the integral change if the function involves cosine instead of sine?

Tip: When evaluating trigonometric integrals, recall identities and properties to simplify integrals wherever possible.