Let R be the region bounded by the given curves. Find the volume of the solid generated when R is revolved about the given line.
xequals​0, yequalsStartRoot x EndRoot​, and yequals​2; about yequals2
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Part 1
A.StartFraction 11 pi Over 4 EndFraction
StartFraction 11 pi Over 4 EndFraction
B.StartFraction 8 pi Over 3 EndFraction
StartFraction 8 pi Over 3 EndFraction
C.StartFraction 10 pi Over 3 EndFraction
StartFraction 10 pi Over 3 EndFraction
D.StartFraction 9 pi Over 4 EndFraction
StartFraction 9 pi Over 4 EndFraction

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.

Let R be the region bounded by the given curves. Find the volume of the solid generated when R is revolved about the given line: x = 0, y = √x, and y = 2; about y = 2.

Learn how to find the volume of the solid generated when the region bounded by x=0, y=√x, and y=2 is revolved around y=2. This calculus problem uses the washer method and explores the application of definite integrals. Suitable for undergraduate calculus students.

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