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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.

Find the volume of the solid generated by rotating the region y = sin(3x), y = 0, x = 0, and x = π/3 about the x-axis.

Learn how to calculate the volume of a solid formed by rotating the region bounded by y = sin(3x), y = 0, and x = 0 to x = π/3 about the x-axis. This solution covers the use of definite integrals and trigonometric identities, suitable for college-level calculus students.

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