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The problem involves finding the volume generated when the shaded region is rotated about the x-axis. Let's outline the steps to solve it:

### Problem Analysis
The given curve is:

\[
y = \frac{4}{(x - 2)^2}
\]

with vertical lines \( x = 3 \) and \( x = 4 \) and horizontal line \( y = 1 \). The shaded region is bounded by these curves and lines, and it will be rotated 360 degrees about the x-axis.

### Step-by-Step Solution
1. **Identify the limits of integration:** The shaded region is between \( x = 3 \) and \( x = 4 \).
  
2. **Volume of Revolution Formula (Disk Method):**
   When a region is rotated about the x-axis, the volume is given by:

   \[
   V = \pi \int_{a}^{b} \left[ f(x) \right]^2 \, dx
   \]

   Here, \( f(x) \) is the curve \( y = \frac{4}{(x-2)^2} \), and \( a = 3 \), \( b = 4 \).

3. **Set up the integral:**

   \[
   V = \pi \int_{3}^{4} \left( \frac{4}{(x-2)^2} \right)^2 \, dx
   \]
   
   Simplifying:

   \[
   V = \pi \int_{3}^{4} \frac{16}{(x-2)^4} \, dx
   \]

4. **Evaluate the integral:**

   Let \( u = x - 2 \), hence \( du = dx \).

   The limits change as follows:
   - When \( x = 3 \), \( u = 3 - 2 = 1 \)
   - When \( x = 4 \), \( u = 4 - 2 = 2 \)

   The integral becomes:

   \[
   V = \pi \int_{1}^{2} \frac{16}{u^4} \, du = \pi \int_{1}^{2} 16u^{-4} \, du
   \]

   Evaluating this integral:

   \[
   \pi \left[ \frac{16u^{-3}}{-3} \right]_{1}^{2}
   \]

   \[
   = \pi \left[ -\frac{16}{3u^3} \right]_{1}^{2}
   \]

   \[
   = \pi \left( -\frac{16}{3 \times 2^3} + \frac{16}{3 \times 1^3} \right)
   \]

   \[
   = \pi \left( -\frac{16}{24} + \frac{16}{3} \right)
   \]

   \[
   = \pi \left( -\frac{2}{3} + \frac{16}{3} \right)
   \]

   \[
   = \pi \left( \frac{14}{3} \right)
   \]

5. **Final volume:**

   \[
   V = \frac{14\pi}{3}
   \]

This is the exact volume of the solid formed when the shaded region is rotated about the x-axis.

Would you like me to go into more details or address any specific part of the solution? Here are some related questions you might consider:

1. What would be the volume if the shaded region was rotated about the y-axis instead of the x-axis?
2. How would the problem change if the limits of integration were from \( x = 2.5 \) to \( x = 4 \)?
3. What is the surface area of the solid formed when the shaded region is rotated about the x-axis?
4. Can we apply the cylindrical shell method for this problem? How would it compare to the disk method?
5. What is the general formula for finding the volume of a solid of revolution?

**Tip:** When setting up integrals for volumes of revolution, carefully choose the method (disk/washer or shell) based on the axis of rotation and the orientation of the function.

Find the exact volume generated when the shaded region is rotated through 360° about the x-axis. The region is bounded by the curve y = 4 / (x - 2)^2, x = 3, x = 4, and y = 1, passing through points A(3, 4) and B(4, 1).

Learn how to calculate the exact volume generated when the region under the curve y = 4 / (x - 2)^2, bounded by x = 3 and x = 4, is rotated around the x-axis using the disk method. This problem involves definite integration and is ideal for calculus students.

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