Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.
y = x2,  x = y2;  about y = 1

To find the volume \( V \) of the solid formed by rotating the region bounded by the curves \( y = x^2 \) and \( x = y^2 \) around the line \( y = 1 \), we will use the **method of cylindrical shells**.

### Step 1: Set up the region
- The curve \( y = x^2 \) is a parabola opening upwards.
- The curve \( x = y^2 \) is a parabola opening sideways.

We need to identify the points of intersection of these two curves. Set \( y = x^2 \) and \( x = y^2 \), which gives:
\[
x = (x^2)^2 \quad \Rightarrow \quad x = x^4.
\]
Solving for \( x \), we have:
\[
x(x^3 - 1) = 0 \quad \Rightarrow \quad x = 0 \quad \text{or} \quad x = 1.
\]
So, the curves intersect at \( x = 0 \) and \( x = 1 \).

### Step 2: Set up the integral
Using the **method of cylindrical shells**, the formula for the volume is:
\[
V = 2\pi \int_{a}^{b} ( \text{radius} ) ( \text{height} ) \, dx.
\]
- The **radius** of a typical shell is the vertical distance from the curve to the line \( y = 1 \), which is \( 1 - y \).
- The **height** of the shell is the horizontal distance between the two curves, which is \( x = y^2 \) and \( y = x^2 \).

We rewrite the problem in terms of \( y \), since the rotation is around the line \( y = 1 \).

### Step 3: Boundaries and limits of integration
- For \( y = 0 \) to \( y = 1 \), the height of the region is given by the curve \( x = y^2 \) and \( y = x^2 \).
  
Thus, we integrate between \( y = 0 \) and \( y = 1 \).

### Step 4: Volume calculation
The integral becomes:
\[
V = 2\pi \int_{0}^{1} (1 - y) \left( y^{1/2} - y^2 \right) \, dy.
\]
Now, we will solve this integral.

### Step 5: Solve the integral

Expanding the expression:
\[
V = 2\pi \int_{0}^{1} \left[ (1 - y)(y^{1/2} - y^2) \right] \, dy
\]
\[
V = 2\pi \int_{0}^{1} \left[ (y^{1/2} - y^2) - y(y^{1/2} - y^2) \right] \, dy
\]
\[
V = 2\pi \int_{0}^{1} \left[ y^{1/2} - y^2 - y^{3/2} + y^3 \right] \, dy.
\]
Now, integrate each term separately:
\[
V = 2\pi \left[ \frac{2}{3}y^{3/2} - \frac{y^3}{3} - \frac{2}{5}y^{5/2} + \frac{y^4}{4} \right]_{0}^{1}.
\]
Evaluating at \( y = 1 \) and \( y = 0 \):
\[
V = 2\pi \left[ \frac{2}{3} - \frac{1}{3} - \frac{2}{5} + \frac{1}{4} \right].
\]
Now simplify:
\[
V = 2\pi \left[ \frac{2}{3} - \frac{1}{3} - \frac{2}{5} + \frac{1}{4} \right] = 2\pi \left[ \frac{1}{3} - \frac{2}{5} + \frac{1}{4} \right].
\]
Find a common denominator (60):
\[
V = 2\pi \left[ \frac{20}{60} - \frac{24}{60} + \frac{15}{60} \right] = 2\pi \left[ \frac{11}{60} \right].
\]
So, the volume is:
\[
V = \frac{22\pi}{60} = \frac{11\pi}{30}.
\]

### Final Answer:
The volume of the solid is \( V = \frac{11\pi}{30} \).

---

Would you like more details or clarifications on any of these steps?

Here are 5 related questions you might want to explore:
1. How do you apply the disk/washer method to solve similar problems?
2. Can this volume be computed using polar coordinates?
3. What is the difference between rotating around \( y = 1 \) and \( y = 0 \)?
4. How would the volume change if we rotated around a vertical line like \( x = 1 \)?
5. How do you find the surface area of the solid formed by this rotation?

**Tip:** When using the method of cylindrical shells, focus on how the radius and height change with respect to the axis of rotation to set up the correct integrals.

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

Learn how to find the volume of a solid formed by rotating the region bounded by y = x^2 and x = y^2 about the line y = 1. This detailed solution uses the method of cylindrical shells and covers essential calculus concepts. Suitable for college-level calculus students.

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