Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.
x = y2,   x = 1 − y2;  
 about x = 4

To find the volume \( V \) of the solid obtained by rotating the region bounded by the curves \( x = y^2 \) and \( x = 1 - y^2 \) about the line \( x = 4 \), we can use the **method of cylindrical shells**.

### Step-by-step solution:

1. **Equations of the curves:**
   - Curve 1: \( x = y^2 \)
   - Curve 2: \( x = 1 - y^2 \)

2. **Determine the limits of integration:**
   To find the points where the curves intersect, set the equations equal to each other:
   \[
   y^2 = 1 - y^2
   \]
   Solving this:
   \[
   2y^2 = 1 \implies y^2 = \frac{1}{2} \implies y = \pm \frac{1}{\sqrt{2}}
   \]
   So the region is bounded between \( y = -\frac{1}{\sqrt{2}} \) and \( y = \frac{1}{\sqrt{2}} \).

3. **Formula for the volume using cylindrical shells:**
   The formula for the volume of a solid of revolution using cylindrical shells is:
   \[
   V = 2\pi \int_{a}^{b} ( \text{radius} ) (\text{height} ) \, dy
   \]
   In this case:
   - The **radius** is the distance from the line of rotation \( x = 4 \) to the curve, so:
     \[
     \text{radius} = 4 - x = 4 - \left( y^2 \right) \quad \text{and} \quad 4 - \left( 1 - y^2 \right)
     \]
   - The **height** is the difference between the two curves:
     \[
     \text{height} = (1 - y^2) - y^2 = 1 - 2y^2
     \]

4. **Set up the integral:**
   \[
   V = 2\pi \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} (4 - y^2)(1 - 2y^2) \, dy
   \]

5. **Simplify the integrand:**
   Expand the terms inside the integral:
   \[
   (4 - y^2)(1 - 2y^2) = 4(1 - 2y^2) - y^2(1 - 2y^2)
   \]
   Simplifying:
   \[
   = 4 - 8y^2 - y^2 + 2y^4 = 4 - 9y^2 + 2y^4
   \]

6. **Integrate:**
   Now, compute the integral:
   \[
   V = 2\pi \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \left( 4 - 9y^2 + 2y^4 \right) \, dy
   \]
   Since the integrand is an even function, we can simplify the integral by doubling the integral from 0 to \( \frac{1}{\sqrt{2}} \):
   \[
   V = 4\pi \int_{0}^{\frac{1}{\sqrt{2}}} \left( 4 - 9y^2 + 2y^4 \right) \, dy
   \]

   Now, calculate each term:
   \[
   \int 4 \, dy = 4y
   \]
   \[
   \int -9y^2 \, dy = -3y^3
   \]
   \[
   \int 2y^4 \, dy = \frac{2y^5}{5}
   \]

7. **Evaluate the definite integrals:**
   Evaluate from \( y = 0 \) to \( y = \frac{1}{\sqrt{2}} \):
   \[
   V = 4\pi \left[ 4y - 3y^3 + \frac{2y^5}{5} \right]_{0}^{\frac{1}{\sqrt{2}}}
   \]
   At \( y = \frac{1}{\sqrt{2}} \):
   \[
   4\left( \frac{1}{\sqrt{2}} \right) = \frac{4}{\sqrt{2}} = 2\sqrt{2}
   \]
   \[
   -3\left( \frac{1}{\sqrt{2}} \right)^3 = -3 \left( \frac{1}{2\sqrt{2}} \right) = -\frac{3}{2\sqrt{2}}
   \]
   \[
   \frac{2 \left( \frac{1}{\sqrt{2}} \right)^5}{5} = \frac{2}{5} \left( \frac{1}{4\sqrt{2}} \right) = \frac{2}{20\sqrt{2}} = \frac{1}{10\sqrt{2}}
   \]

8. **Final result:**
   The volume \( V \) of the solid is:
   \[
   V = 4\pi \left( 2\sqrt{2} - \frac{3}{2\sqrt{2}} + \frac{1}{10\sqrt{2}} \right)
   \]

If you'd like the numerical value or further simplification of the volume, let me know!

---

Would you like more details or have any specific questions?

Here are 5 related questions to explore further:
1. How would the method change if the axis of rotation were a vertical line?
2. What if the curves were rotated about the \( y \)-axis?
3. How do you compute the volume using the disk or washer method for this problem?
4. Can you apply the cylindrical shell method for rotation around other horizontal lines, such as \( x = 2 \)?
5. What happens to the volume if the curves are shifted horizontally?

**Tip**: For problems involving rotation about non-axis lines, always adjust the radius as the distance from the curve to the line of rotation!

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. x = y^2, x = 1 − y^2; about x = 4.

Calculate the volume of the solid formed by rotating the region bounded by x = y^2 and x = 1 - y^2 about the line x = 4 using the cylindrical shells method. This problem involves the application of calculus concepts such as definite integrals and is suitable for students in Grades 11-12.

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