To find the volume of the solid generated by revolving the region bounded by the curve $y = 2x$, $y = 0$, and $x = 3$ about the line $y = -2$ using the shell method, we proceed as follows:

Step 1: Understanding the Setup

The region of interest is between:

• The line $y = 2x$ (from $x = 0$ to $x = 3$),
• The $y = 0$ (x-axis),
• And the vertical line $x = 3$.

We are rotating this region about the line $y = -2$, which is a horizontal line below the x-axis.

Step 2: Formula for Shell Method

The formula for the volume using the shell method is:

$V = 2\pi \int_{a}^{b} ( \text{radius}(x) \cdot \text{height}(x) ) \, dx$

Where:

• Radius is the distance from the line of rotation (which is $y = -2$) to the shell.
• Height is the function representing the height of the shell.

Step 3: Define the Radius and Height

• Height of the shell: The height at any $x$ is the distance between $y = 2x$ and $y = 0$, which is just $2x$ (the given function).

• Radius of the shell: The distance from a point on the curve to the line $y = -2$. This distance is $y + 2$, where $y = 2x$. Therefore, the radius is $2x + 2$.

Step 4: Set Up the Integral

The integral is set up from $x = 0$ to $x = 3$, since these are the limits of the region. The volume is:

$V = 2\pi \int_{0}^{3} ( (2x + 2) \cdot 2x ) \, dx$

Step 5: Simplify the Integral

Expand the integrand:

$(2x + 2)(2x) = 4x^2 + 4x$

Thus, the volume integral becomes:

$V = 2\pi \int_{0}^{3} (4x^2 + 4x) \, dx$

Step 6: Integrate

Now, we integrate each term:

$\int_{0}^{3} 4x^2 \, dx = \frac{4}{3}x^3 \Big|_0^3 = \frac{4}{3}(27) - 0 = 36$

$\int_{0}^{3} 4x \, dx = 2x^2 \Big|_0^3 = 2(9) - 0 = 18$

Step 7: Compute the Volume

Now, add the results of the integrals:

$V = 2\pi (36 + 18) = 2\pi (54) = 108\pi$

Thus, the volume of the solid is:

$\boxed{108\pi}$

Would you like more details on any part of this process?

Here are some related questions you could explore:

1. How does the shell method differ from the disk/washer method?
2. What happens if the axis of rotation is shifted further down?
3. How would the volume change if the region were rotated about the x-axis?
4. What if the function was $y = x^2$ instead of $y = 2x$?
5. Can the shell method be applied when revolving around vertical lines?

Tip: When using the shell method, always check if the radius is the distance from the curve to the axis of rotation, not just the function itself.