lets find the volume of a solid revolving about line x=1. IT is bounded by curves y=x^2+1, x=0, x=1, y=0

To find the volume of the solid formed by revolving the region bounded by the curves \( y = x^2 + 1 \), \( x = 0 \), \( x = 1 \), and \( y = 0 \) about the line \( x = 1 \), we can use the **method of cylindrical shells**. This method is particularly useful when revolving a region around a vertical line that is not one of the axes.

### Step-by-Step Solution:

1. **Region Boundaries**:
   - \( y = x^2 + 1 \) (upper boundary)
   - \( y = 0 \) (lower boundary)
   - \( x = 0 \) (left boundary)
   - \( x = 1 \) (right boundary)

2. **Cylindrical Shells Method**:
   When rotating about the vertical line \( x = 1 \), the formula for the volume of a cylindrical shell is:

   \[
   V = 2\pi \int_{a}^{b} ( \text{radius} ) ( \text{height} ) \, dx
   \]

   - **Radius**: The distance from the line of revolution \( x = 1 \) to a point \( x \) is \( |1 - x| = 1 - x \).
   - **Height**: The height of the shell is the function \( y = x^2 + 1 \).

   So, the volume integral becomes:
   \[
   V = 2\pi \int_{0}^{1} (1 - x)(x^2 + 1) \, dx
   \]

3. **Expand the integrand**:
   \[
   (1 - x)(x^2 + 1) = (1 \cdot x^2) + (1 \cdot 1) - (x \cdot x^2) - (x \cdot 1)
   \]
   Simplifying:
   \[
   = x^2 + 1 - x^3 - x
   \]

   Thus, the integral becomes:
   \[
   V = 2\pi \int_{0}^{1} (x^2 + 1 - x^3 - x) \, dx
   \]

4. **Integrate each term**:
   \[
   \int (x^2 + 1 - x^3 - x) \, dx = \left[\frac{x^3}{3} + x - \frac{x^4}{4} - \frac{x^2}{2}\right]_{0}^{1}
   \]

5. **Evaluate the integral** at the bounds \( 0 \) and \( 1 \):
   - At \( x = 1 \):
     \[
     \frac{1^3}{3} + 1 - \frac{1^4}{4} - \frac{1^2}{2} = \frac{1}{3} + 1 - \frac{1}{4} - \frac{1}{2}
     \]
     Combine the fractions:
     \[
     = \frac{1}{3} + 1 - \frac{1}{4} - \frac{1}{2} = \frac{4}{12} + \frac{12}{12} - \frac{3}{12} - \frac{6}{12}
     \]
     Simplifying:
     \[
     = \frac{4 + 12 - 3 - 6}{12} = \frac{7}{12}
     \]

   - At \( x = 0 \), all terms are 0.

6. **Multiply by \( 2\pi \)**:
   \[
   V = 2\pi \times \frac{7}{12} = \frac{7\pi}{6}
   \]

### Final Answer:
The volume of the solid is \( \frac{7\pi}{6} \).

---

Would you like more details or have any further questions? Here are some related problems for you to explore:

1. Find the volume if the region is revolved around the line \( x = 2 \).
2. What is the volume when the same region is revolved around the y-axis?
3. Find the centroid of the region before revolving.
4. Use the disk method to compute the volume revolving around the x-axis.
5. How does the volume change if the upper boundary is \( y = 2x^2 + 1 \)?

**Tip:** The method of cylindrical shells is generally easier for problems involving revolutions around vertical lines.

Calculate the volume of a solid formed by revolving the region bounded by y=x^2+1, x=0, x=1, and y=0 about the line x=1 using the Cylindrical Shells method. This step-by-step guide includes integration techniques and is suitable for college-level calculus students.

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