Use the shell method to find the volume of the solid generated by revolving the region by y=2x ,y=0,x=3 about the line y=-2

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.

Use the shell method to find the volume of the solid generated by revolving the region by y=2x, y=0, x=3 about the line y=-2.

Learn how to find the volume of the solid generated by revolving the region y=2x, y=0, and x=3 about the line y=-2 using the shell method. This step-by-step solution covers key calculus concepts and shell method formulas, ideal for advanced calculus students in Grades 11-12.

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