Use the shell method to find the volume of the solid generated by revolving the regions bounded by the curves and lines about the​ x-axis.
yequalsStartRoot x EndRoot​,   yequals​0,   yequalsx minus 2

To find the volume of the solid generated by revolving the region bounded by the curves \( y = \sqrt{x} \), \( y = 0 \), and \( y = x - 2 \) about the x-axis using the shell method, we follow these steps:

### Step 1: Sketch the Region
First, sketch the region bounded by the curves:
- \( y = \sqrt{x} \): This is a parabola opening to the right.
- \( y = 0 \): This is the x-axis.
- \( y = x - 2 \): This is a straight line with a slope of 1 and y-intercept of -2.

### Step 2: Find the Points of Intersection
To determine the bounds of integration, find the points of intersection between \( y = \sqrt{x} \) and \( y = x - 2 \).

Set \( \sqrt{x} = x - 2 \) and solve for \( x \).

1. Square both sides:
   \[
   x = (x - 2)^2
   \]
2. Expand and simplify:
   \[
   x = x^2 - 4x + 4
   \]
   \[
   0 = x^2 - 5x + 4
   \]
3. Factor the quadratic equation:
   \[
   0 = (x - 4)(x - 1)
   \]
   \[
   x = 4 \quad \text{or} \quad x = 1
   \]
So, the points of intersection occur at \( x = 1 \) and \( x = 4 \).

### Step 3: Set Up the Shell Method Integral
The shell method formula for rotating around the x-axis is:
\[
V = 2\pi \int_{a}^{b} y \cdot \text{(radius)} \cdot \text{(height)} \, dy
\]
Here, the radius of each shell is \( y \), and the height is given by the difference in x-values (which are functions of y).

Solve for \( x \) in terms of \( y \) from each equation:
- For \( y = \sqrt{x} \), \( x = y^2 \).
- For \( y = x - 2 \), \( x = y + 2 \).

So, the height of the shell is \( (y + 2) - y^2 \).

### Step 4: Integrate
The bounds for \( y \) go from \( y = 0 \) (lower bound) to the y-value at \( x = 4 \), which is \( y = 2 \).

Set up the integral:
\[
V = 2\pi \int_{0}^{2} y[(y + 2) - y^2] \, dy
\]

Simplify the integrand:
\[
V = 2\pi \int_{0}^{2} (y^2 + 2y - y^3) \, dy
\]

### Step 5: Solve the Integral
Now, integrate term by term:
\[
V = 2\pi \left[\frac{y^3}{3} + y^2 - \frac{y^4}{4}\right]_{0}^{2}
\]

Evaluate at the bounds:
\[
V = 2\pi \left[\left(\frac{2^3}{3} + 2^2 - \frac{2^4}{4}\right) - \left(\frac{0^3}{3} + 0^2 - \frac{0^4}{4}\right)\right]
\]
\[
V = 2\pi \left[\left(\frac{8}{3} + 4 - 4\right) - 0\right]
\]
\[
V = 2\pi \times \frac{8}{3} = \frac{16\pi}{3}
\]

### Final Answer
The volume of the solid generated by revolving the given region around the x-axis is:
\[
\boxed{\frac{16\pi}{3}}
\]

Would you like more details or have any questions?

---

Here are five related questions to explore further:
1. How does the method change if the region is revolved around the y-axis instead of the x-axis?
2. What is the significance of choosing the shell method over the disk method?
3. Can you solve for the volume using the disk method instead? How does the setup differ?
4. How would the volume calculation change if one of the curves were \( y = \sqrt{2x} \) instead?
5. What are the general steps to follow when identifying the boundaries of the region for rotation?

**Tip:** Always double-check the points of intersection and the orientation of the curves to ensure correct limits of integration.

Use the shell method to find the volume of the solid generated by revolving the regions bounded by the curves and lines about the x-axis. y = sqrt(x), y = 0, y = x - 2.

Calculate the volume of a solid generated by revolving the region bounded by y = sqrt(x), y = 0, and y = x - 2 around the x-axis using the shell method. This detailed solution explores key concepts in calculus, including the shell method and the fundamental theorem of calculus. Ideal for college-level calculus students.

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