Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.
y = x,  y = sqrtx;  about x = 2

To find the volume \( V \) of the solid obtained by rotating the region bounded by the curves \( y = x \) and \( y = \sqrt{x} \) about the line \( x = 2 \), we can use the **method of cylindrical shells**.

### Step-by-Step Solution:

1. **Identify the region and the curves**:
   - The curves are \( y = x \) and \( y = \sqrt{x} \).
   - The region is bounded by these curves.

2. **Find the intersection points**:
   - Set \( y = x \) equal to \( y = \sqrt{x} \) to find the intersection points:
     \[
     x = \sqrt{x}
     \]
     Squaring both sides:
     \[
     x^2 = x
     \]
     Rearrange to:
     \[
     x^2 - x = 0
     \]
     Factor:
     \[
     x(x - 1) = 0
     \]
     So, \( x = 0 \) and \( x = 1 \).

     The region of integration is from \( x = 0 \) to \( x = 1 \).

3. **Set up the volume integral using the method of cylindrical shells**:
   - The formula for the volume using cylindrical shells is:
     \[
     V = 2\pi \int_{a}^{b} (r_{\text{shell}})(h_{\text{shell}}) \, dx
     \]
   - Here, the radius of the shell is the distance from the curve to the line \( x = 2 \):
     \[
     r_{\text{shell}} = 2 - x
     \]
   - The height of the shell is the difference between the upper and lower curves:
     \[
     h_{\text{shell}} = \sqrt{x} - x
     \]

   Therefore, the integral becomes:
   \[
   V = 2\pi \int_{0}^{1} (2 - x)(\sqrt{x} - x) \, dx
   \]

4. **Expand and simplify the integrand**:
   \[
   V = 2\pi \int_{0}^{1} \left[(2 - x)\sqrt{x} - (2 - x)x\right] \, dx
   \]
   Distribute the terms:
   \[
   V = 2\pi \int_{0}^{1} \left[2\sqrt{x} - x\sqrt{x} - 2x + x^2\right] \, dx
   \]

5. **Integrate term by term**:
   - For \( 2\sqrt{x} \), use \( \int \sqrt{x} \, dx = \frac{2}{3}x^{3/2} \):
     \[
     2\int \sqrt{x} \, dx = \frac{4}{3}x^{3/2}
     \]
   - For \( -x\sqrt{x} \), use \( \int x^{3/2} \, dx = \frac{2}{5}x^{5/2} \):
     \[
     -\int x\sqrt{x} \, dx = -\frac{2}{5}x^{5/2}
     \]
   - For \( -2x \), use \( \int x \, dx = \frac{1}{2}x^2 \):
     \[
     -2\int x \, dx = -x^2
     \]
   - For \( x^2 \), use \( \int x^2 \, dx = \frac{1}{3}x^3 \):
     \[
     \int x^2 \, dx = \frac{1}{3}x^3
     \]

6. **Evaluate the definite integral**:
   \[
   V = 2\pi \left[\left(\frac{4}{3}x^{3/2} - \frac{2}{5}x^{5/2} - x^2 + \frac{1}{3}x^3\right)\bigg|_{0}^{1}\right]
   \]
   At \( x = 1 \):
   \[
   \frac{4}{3}(1) - \frac{2}{5}(1) - (1) + \frac{1}{3}(1) = \frac{4}{3} - \frac{2}{5} - 1 + \frac{1}{3}
   \]
   At \( x = 0 \), the integral evaluates to 0.

   Therefore:
   \[
   V = 2\pi \left(\frac{4}{3} + \frac{1}{3} - 1 - \frac{2}{5}\right) = 2\pi \left(\frac{5}{3} - \frac{2}{5}\right) = 2\pi \left(\frac{25}{15} - \frac{6}{15}\right) = 2\pi \left(\frac{19}{15}\right)
   \]
   \[
   V = \frac{38\pi}{15}
   \]

### Final Answer:
The volume \( V \) of the solid is \( \boxed{\frac{38\pi}{15}} \) cubic units.

Would you like further details on any of the steps? Here are five related questions:

1. How do you set up the volume integral using the washer method for this problem?
2. What is the difference between the method of cylindrical shells and the disk/washer method?
3. How do you visualize the solid formed by rotating the region around the line \( x = 2 \)?
4. Can we use a different axis of rotation and how does that affect the integral setup?
5. How would the problem change if the region was rotated around the y-axis instead?

**Tip**: When rotating around vertical lines (not the y-axis), the method of cylindrical shells is often more convenient than the disk/washer method.

Learn how to find the volume V of the solid obtained by rotating the region bounded by y = x and y = sqrt(x) around the line x = 2. Step-by-step solution using the method of cylindrical shells with detailed integration and calculation. Suitable for advanced mathematics enthusiasts.

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