I need to find the volume of the region bounded by y=x^2+1, x=0, x=1, y=0, revolving around y=2. How would I slice the region and set up the problem

To find the volume of the region bounded by \( y = x^2 + 1 \), \( x = 0 \), \( x = 1 \), and \( y = 0 \), revolving around the line \( y = 2 \), we can use the **method of cylindrical shells** or **washer method**. Here, I will guide you through the **washer method** since the region is revolving around a horizontal line that is not the axis of rotation.

### Step-by-step approach:

1. **Understand the bounds and shape**:
   - The curve is given by \( y = x^2 + 1 \), so for \( x = 0 \), we have \( y = 1 \), and for \( x = 1 \), we have \( y = 2 \). 
   - We're rotating around \( y = 2 \), and our limits for \( x \) are from 0 to 1.

2. **Washer method setup**:
   - The general formula for the volume of a solid of revolution using the washer method is:
     \[
     V = \pi \int_{a}^{b} \left[ R_{\text{outer}}^2 - R_{\text{inner}}^2 \right] dx
     \]
     Here, \( R_{\text{outer}} \) and \( R_{\text{inner}} \) represent the distances of the region's boundaries from the axis of rotation (which is \( y = 2 \)).

3. **Determine radii**:
   - The **outer radius** is the distance from the axis \( y = 2 \) to the line \( y = 1 \) (which corresponds to \( x = 0 \)): this is constant and equals \( R_{\text{outer}} = 2 - 1 = 1 \).
   - The **inner radius** is the distance from the axis \( y = 2 \) to the curve \( y = x^2 + 1 \), so \( R_{\text{inner}} = 2 - (x^2 + 1) = 1 - x^2 \).

4. **Set up the integral**:
   We now set up the integral to calculate the volume. The volume is:
   \[
   V = \pi \int_0^1 \left( 1^2 - (1 - x^2)^2 \right) dx
   \]
   Expanding \( (1 - x^2)^2 \):
   \[
   (1 - x^2)^2 = 1 - 2x^2 + x^4
   \]
   So the integrand becomes:
   \[
   1 - (1 - 2x^2 + x^4) = 2x^2 - x^4
   \]
   Now the volume is:
   \[
   V = \pi \int_0^1 (2x^2 - x^4) dx
   \]

5. **Evaluate the integral**:
   \[
   V = \pi \left[ \frac{2x^3}{3} - \frac{x^5}{5} \right]_0^1
   \]
   Substituting the limits:
   \[
   V = \pi \left( \frac{2(1)^3}{3} - \frac{(1)^5}{5} \right) = \pi \left( \frac{2}{3} - \frac{1}{5} \right)
   \]
   Simplifying:
   \[
   V = \pi \left( \frac{10}{15} - \frac{3}{15} \right) = \pi \times \frac{7}{15}
   \]
   Therefore, the volume is:
   \[
   V = \frac{7\pi}{15}
   \]

### Final Answer:
The volume of the solid generated by revolving the given region around \( y = 2 \) is:
\[
V = \frac{7\pi}{15}
\]

Would you like further details or have any questions?

---

Here are five related questions to deepen your understanding:

1. How would the integral setup change if the region was revolved around \( y = 3 \) instead of \( y = 2 \)?
2. What if the curve \( y = x^2 + 1 \) was revolved around the \( x \)-axis instead of \( y = 2 \)?
3. How would you set up the volume calculation using the cylindrical shell method for this problem?
4. What happens if you extend the region to \( x = 2 \)? How does the volume change?
5. How would the setup differ if the lower bound for \( y \) was \( y = 1.5 \) instead of \( y = 1 \)?

### Tip:
When choosing between the washer and shell methods, think about the axis of rotation. If the axis is parallel to the axis of integration (here, a horizontal line and integrating in terms of \( x \)), the washer method is often simpler.

I need to find the volume of the region bounded by y=x^2+1, x=0, x=1, y=0, revolving around y=2. How would I slice the region and set up the problem?

Learn how to calculate the volume of the region bounded by y=x^2+1, x=0, x=1, y=0, revolving around y=2 using the washer method. This problem involves key calculus concepts, including integration and solid of revolution techniques. Suitable for undergraduate calculus students.

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