The problem asks for the area of the surface generated by revolving the curve $y = \frac{1}{3}x^3$ from $x = 0$ to $x = 3$ around the x-axis. To calculate the surface area of revolution around the x-axis, we use the formula for surface area $S$ of a curve $y = f(x)$ rotated around the x-axis:

$S = 2\pi \int_{a}^{b} y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx$

Step 1: Define the function and its derivative

Given $y = \frac{1}{3}x^3$, let's first find the derivative of $y$ with respect to $x$.

$\frac{dy}{dx} = x^2$

Step 2: Set up the integral

Now substitute $y = \frac{1}{3}x^3$ and $\frac{dy}{dx} = x^2$ into the formula:

$S = 2\pi \int_{0}^{3} \left( \frac{1}{3}x^3 \right) \sqrt{1 + (x^2)^2} \, dx$

This simplifies to:

$S = 2\pi \int_{0}^{3} \frac{1}{3}x^3 \sqrt{1 + x^4} \, dx$

Step 3: Evaluate the integral

This integral does not have an elementary antiderivative, so we will approximate the value numerically. Let's calculate it.The surface area generated by revolving the curve $y = \frac{1}{3}x^3$ from $x = 0$ to $x = 3$ around the x-axis is approximately $258.85$ square units.

Would you like further details or have any questions?

Here are 5 related questions to expand on this concept:

1. How would the surface area change if we rotated around the y-axis instead?
2. Can we compute the surface area for a different range, say from $x = 1$ to $x = 3$?
3. What would happen to the surface area if the function were $y = \frac{1}{2}x^2$?
4. How can we find the volume generated by rotating this curve around the x-axis?
5. How does the surface area formula differ if we rotate around a line other than the x-axis?

Tip: When setting up surface integrals, carefully handle the derivative term, as it significantly influences the final result!