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.