We are tasked with finding the volume of the solid generated when the region bounded by $x = 0$, $y = \sqrt{x}$, and $y = 2$ is revolved around the line $y = 2$.

Step 1: Identify the region $R$

The curves given are:

• $x = 0$: The vertical line on the $y$-axis.
• $y = \sqrt{x}$: This is a parabola that opens to the right.
• $y = 2$: A horizontal line.

We need to rotate this region around the line $y = 2$, and we'll use the disk/washer method for this.

Step 2: Set up the washer method

The general formula for the volume using the washer method is:

$V = \pi \int_{a}^{b} \left( R_{\text{outer}}(x)^2 - R_{\text{inner}}(x)^2 \right) dx$

Here, the "outer radius" is the distance from the line $y = 2$ to the $x$-axis (the height of the line $y = 2$), and the "inner radius" is the distance from $y = 2$ to the curve $y = \sqrt{x}$.

Step 3: Calculate the outer and inner radii

• The outer radius is the distance from the line $y = 2$ to $y = 0$, which is simply 2.

• The inner radius is the distance from the line $y = 2$ to the curve $y = \sqrt{x}$, which is $2 - \sqrt{x}$.

Step 4: Set up the integral

We need to revolve the region between $x = 0$ and the point where the curve $y = \sqrt{x}$ intersects $y = 2$. Solving $y = \sqrt{x} = 2$, we get $x = 4$.

So, the volume integral becomes:

$V = \pi \int_{0}^{4} \left(2^2 - (2 - \sqrt{x})^2\right) dx$

Step 5: Simplify the integrand

First, expand $(2 - \sqrt{x})^2$:

$(2 - \sqrt{x})^2 = 4 - 4\sqrt{x} + x$

Thus, the integrand becomes:

$4 - (4 - 4\sqrt{x} + x) = 4 - 4 + 4\sqrt{x} - x = 4\sqrt{x} - x$

Step 6: Compute the integral

Now, we compute the integral:

$V = \pi \int_{0}^{4} (4\sqrt{x} - x) dx$

Break it up:

$V = \pi \left[ \int_{0}^{4} 4\sqrt{x} dx - \int_{0}^{4} x dx \right]$

For $\int 4\sqrt{x} dx$, we use $\int 4x^{1/2} dx = \frac{4}{3} x^{3/2}$:

$\int_{0}^{4} 4\sqrt{x} dx = \frac{4}{3} (4)^{3/2} - \frac{4}{3} (0)^{3/2} = \frac{4}{3} \times 8 = \frac{32}{3}$

For $\int x dx = \frac{x^2}{2}$:

$\int_{0}^{4} x dx = \frac{4^2}{2} - \frac{0^2}{2} = \frac{16}{2} = 8$

Step 7: Final calculation

Now, subtract the two integrals:

$V = \pi \left( \frac{32}{3} - 8 \right) = \pi \left( \frac{32}{3} - \frac{24}{3} \right) = \pi \times \frac{8}{3} = \frac{8\pi}{3}$

Final Answer:

The volume of the solid is:

$\boxed{\frac{8\pi}{3}}$

This corresponds to option B.

Would you like more details or have any questions about this? Here are 5 related questions to expand your understanding:

1. How does the washer method differ from the shell method in calculating volumes of revolution?
2. Can the solid's volume change if we revolve the region around a different axis, such as the $x$-axis?
3. What happens if the region is revolved around the line $y = 3$ instead of $y = 2$?
4. How would the problem change if the region were bounded by different curves, such as $y = \ln(x)$?
5. Can you use a different integration method, like cylindrical shells, to solve this problem?

Tip: When using the washer method, make sure you correctly identify the outer and inner radii, especially when the axis of rotation is not one of the coordinate axes.