To solve the given problem using the shell method, we will find the volume of the solid generated by revolving the region bounded by $y = 2x$, $y = -\frac{x}{2}$, and $x = 3$ about the y-axis.

The shell method involves integrating along the axis perpendicular to the axis of revolution. In this case, since we are revolving around the y-axis, we will integrate with respect to $x$.

1. Determine the height and radius of the shell:

   • The height of the shell at a given $x$ is the difference between the two curves, $y = 2x$ and $y = -\frac{x}{2}$: $\text{Height} = 2x - \left( -\frac{x}{2} \right) = 2x + \frac{x}{2} = \frac{5x}{2}$
   • The radius of the shell is the distance from the y-axis, which is simply $x$.

2. Set up the integral:

   • The volume $V$ using the shell method is given by: $V = 2\pi \int_{a}^{b} (\text{radius})(\text{height}) \, dx$ Here, $a = 0$ and $b = 3$.

     Substituting the radius and height, we get: $V = 2\pi \int_{0}^{3} x \cdot \frac{5x}{2} \, dx = 2\pi \int_{0}^{3} \frac{5x^2}{2} \, dx$

3. Simplify the integral: $V = 2\pi \cdot \frac{5}{2} \int_{0}^{3} x^2 \, dx = 5\pi \int_{0}^{3} x^2 \, dx$

4. Evaluate the integral: $\int_{0}^{3} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{3} = \left( \frac{3^3}{3} - \frac{0^3}{3} \right) = \left( \frac{27}{3} - 0 \right) = 9$ Therefore, $V = 5\pi \cdot 9 = 45\pi$

The volume of the solid generated by revolving the given region about the y-axis is $45\pi$ cubic units.

Would you like more details or have any questions?

Here are some related questions that you might find useful:

1. How does the shell method compare to the disk method for finding volumes of solids of revolution?
2. Can the shell method be used for regions revolved around the x-axis, and if so, how?
3. What are some common mistakes to avoid when using the shell method?
4. How can you verify your answer when using the shell method?
5. How do the limits of integration change if the region is revolved around a different axis?

Tip: Always sketch the region and the axis of revolution to better visualize the problem and avoid mistakes in setting up the integral.