To solve this problem using the shell method, we'll follow these steps:

1. Define the Shell Method Formula

The shell method formula for the volume of a solid of revolution about the $y$-axis is: $V = 2\pi \int_{a}^{b} (radius)(height) \, dx,$ where:

• radius is the distance from the shell to the axis of rotation, which is $x$.
• height is the function describing the height of the shell.

2. Identify the Radius and Height

For this problem:

• radius is $x$, because we are rotating around the $y$-axis.
• height is the difference between the upper function and the lower function: $\text{height} = (2x) - \left(-\frac{x}{2}\right) = \frac{5x}{2}.$

3. Set Up the Integral

The region is bounded by $x = 0$ to $x = 3$. The volume integral becomes: $V = 2\pi \int_{0}^{3} x \left(\frac{5x}{2}\right) \, dx = 2\pi \int_{0}^{3} \frac{5x^2}{2} \, dx.$

4. Integrate

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

Thus, the volume of the solid generated is $45\pi$ cubic units.

Would you like more details or have any questions about this solution?

Related Questions

1. How would the solution change if the region was revolved around the x-axis instead?
2. What if the equations for the boundaries were different, such as $y = 3x$ and $y = -x/3$?
3. Can the disk method be used to solve this problem? Why or why not?
4. How do you determine which method (shell or disk) to use for volume of revolution problems?
5. What are the implications of revolving around a different line, such as $x = -2$?

Tip

When setting up integrals for volume using the shell or disk methods, always carefully identify the radius and height in relation to the axis of rotation to avoid mistakes.