If the region enclosed by the straight lines y= x x= 1 and the y= 0 is revolved about the x=1, compute the volume of the solid so generated.

To compute the volume of the solid generated by revolving the region enclosed by the lines \( y = x \), \( x = 1 \), and \( y = 0 \) about the line \( x = 1 \), we can use the method of cylindrical shells.

### Step 1: Identify the region and setup
The region is bounded by:
- The curve \( y = x \)
- The vertical line \( x = 1 \)
- The horizontal line \( y = 0 \) (which is the \( x \)-axis)

When revolving around the line \( x = 1 \), the cylindrical shells method is more convenient.

### Step 2: Cylindrical Shells Formula
The volume \( V \) of the solid of revolution can be calculated using the cylindrical shells method:
\[
V = 2\pi \int_{a}^{b} ( \text{radius} ) \times (\text{height}) \, dx
\]
where:
- **Radius** is the distance from a point on the curve to the axis of rotation.
- **Height** is the function value \( y = x \).

Here, the axis of rotation is \( x = 1 \), so the radius is \( 1 - x \) and the height is \( y = x \). The limits of integration are from \( x = 0 \) to \( x = 1 \).

### Step 3: Set up the integral
\[
V = 2\pi \int_{0}^{1} (1 - x) \times x \, dx
\]

### Step 4: Compute the integral
Expand the integrand:
\[
V = 2\pi \int_{0}^{1} (x - x^2) \, dx
\]

Now, integrate term by term:
\[
\int (x - x^2) \, dx = \frac{x^2}{2} - \frac{x^3}{3}
\]

Evaluate this from 0 to 1:
\[
V = 2\pi \left[ \left( \frac{x^2}{2} - \frac{x^3}{3} \right) \right]_{0}^{1}
\]
\[
V = 2\pi \left( \frac{1^2}{2} - \frac{1^3}{3} - \left( \frac{0^2}{2} - \frac{0^3}{3} \right) \right)
\]
\[
V = 2\pi \left( \frac{1}{2} - \frac{1}{3} \right)
\]

Simplify the expression:
\[
V = 2\pi \left( \frac{3}{6} - \frac{2}{6} \right)
\]
\[
V = 2\pi \times \frac{1}{6}
\]
\[
V = \frac{2\pi}{6} = \frac{\pi}{3}
\]

### Final Answer:
The volume of the solid generated by revolving the given region about the line \( x = 1 \) is \( \frac{\pi}{3} \) cubic units.

Would you like further details or have any questions?

Here are 5 related questions you might find interesting:
1. How would the volume change if the region was revolved around the line \( x = 2 \)?
2. What would be the volume if the region was revolved around the \( y \)-axis instead?
3. How does the method of washers compare to cylindrical shells in this case?
4. Can you calculate the surface area of the solid generated by this revolution?
5. What if the curve was \( y = x^2 \) instead of \( y = x \)? How would that affect the volume?

**Tip:** When using the method of cylindrical shells, remember to carefully determine the radius and height relative to the axis of rotation to set up the integral correctly.

Learn how to compute the volume of a solid generated by revolving the region bounded by the lines y = x, x = 1, and y = 0 about the line x = 1 using the cylindrical shells method. Step-by-step solution with integral calculus. Suitable for advanced high school students.

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