Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line, using disk/washer method. Sketch the region, the solid, and a typical disk or washer. 𝑥 = 0, 𝑦 = √𝑥, and 𝑦 = 2 about the line 𝑥 = 4

To solve this problem using the disk/washer method, we will follow the steps outlined below:

### Step 1: Understand the region to be rotated.
The region is bounded by:
- \( x = 0 \) (the y-axis),
- \( y = \sqrt{x} \) (the curve),
- \( y = 2 \) (a horizontal line).

We will rotate this region about the vertical line \( x = 4 \).

### Step 2: Express the curves in terms of \( y \).
Since we are rotating around a vertical line, we need to re-express the equation of the curve \( y = \sqrt{x} \) in terms of \( y \). Solving for \( x \), we get:
\[
x = y^2
\]
Thus, the region is bounded by \( x = 0 \), \( x = y^2 \), and \( y = 2 \).

### Step 3: Set up the washer method.
The volume of the solid is obtained by integrating the area of washers (rings). Each washer has:
- **Outer radius**: the distance from the axis of rotation (\( x = 4 \)) to the outer boundary of the region, which is \( x = 0 \) (the y-axis). The outer radius is \( R(y) = 4 - 0 = 4 \).
- **Inner radius**: the distance from \( x = 4 \) to the curve \( x = y^2 \), which is \( R_{\text{inner}}(y) = 4 - y^2 \).

### Step 4: Set up the integral for the volume.
The formula for the volume using the washer method is:
\[
V = \pi \int_{y_{\text{min}}}^{y_{\text{max}}} \left[ R_{\text{outer}}^2(y) - R_{\text{inner}}^2(y) \right] \, dy
\]
Substitute the limits of integration from \( y = 0 \) to \( y = 2 \), and the radii:
\[
V = \pi \int_0^2 \left[ (4^2) - (4 - y^2)^2 \right] \, dy
\]
Simplify the expressions inside the integral:
\[
V = \pi \int_0^2 \left[ 16 - (16 - 8y^2 + y^4) \right] \, dy
\]
\[
V = \pi \int_0^2 \left[ 16 - 16 + 8y^2 - y^4 \right] \, dy
\]
\[
V = \pi \int_0^2 \left[ 8y^2 - y^4 \right] \, dy
\]

### Step 5: Compute the integral.
Now, integrate term by term:
\[
V = \pi \left[ \frac{8y^3}{3} - \frac{y^5}{5} \right]_0^2
\]
Evaluate at \( y = 2 \):
\[
V = \pi \left[ \frac{8(2)^3}{3} - \frac{(2)^5}{5} \right]
\]
\[
V = \pi \left[ \frac{64}{3} - \frac{32}{5} \right]
\]
Find a common denominator:
\[
V = \pi \left[ \frac{320}{15} - \frac{96}{15} \right]
\]
\[
V = \pi \left[ \frac{224}{15} \right]
\]

Thus, the volume is:
\[
V = \frac{224\pi}{15}
\]

### Step 6: Sketching the region and the solid.
1. **Region**: The region is the area bounded by \( x = 0 \), \( y = \sqrt{x} \), and \( y = 2 \).
2. **Solid**: The solid is obtained by rotating this region about the line \( x = 4 \), creating a solid that resembles a hollow cylinder with varying inner radii.
3. **Typical washer**: A typical washer has an outer radius of \( 4 \) and an inner radius of \( 4 - y^2 \).

Would you like to see the sketch, or is there anything you would like more details on?

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### Related Questions:
1. What happens to the volume if we rotate about the line \( x = -4 \) instead?
2. How would the method change if the region were rotated about the \( y \)-axis?
3. Can you use cylindrical shells instead of the washer method for this problem?
4. How does the volume change if we replace the curve \( y = \sqrt{x} \) with \( y = x^2 \)?
5. What if the boundary \( y = 2 \) was replaced by \( y = 1 \)?

### Tip:
Always remember to carefully set up the limits of integration based on the axis of rotation, whether vertical or horizontal.

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line, using disk/washer method. Sketch the region, the solid, and a typical disk or washer. x = 0, y = √x, and y = 2 about the line x = 4.

This page explains how to find the volume of the solid formed by rotating the region bounded by x = 0, y = √x, and y = 2 around the line x = 4 using the washer method. It covers integral setup, radius calculations, and step-by-step solution, suitable for advanced high school or undergraduate calculus students.

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