Find the volume of the solid obtained by rotating the region about the x-axis given the region enclosed by y=x and y=x^2. Now, Find the volume of the solid obtained by rotating the region about the y-axis enclosed by the same region.

### 1. Volume of the solid obtained by rotating the region about the **x-axis**:

The region is enclosed by the curves \( y = x \) and \( y = x^2 \), and we need to find the volume of the solid obtained by rotating this region around the x-axis. To do so, we'll use the **disk method**.

- **Limits of integration**: The two curves intersect when \( y = x^2 \) and \( y = x \). Setting \( x = x^2 \) gives:
  \[
  x^2 - x = 0 \quad \Rightarrow \quad x(x - 1) = 0
  \]
  So, the region is bounded between \( x = 0 \) and \( x = 1 \).

- **Volume formula using disk method**: For rotation about the x-axis, the volume \( V \) is given by:
  \[
  V = \pi \int_{a}^{b} \left[ R_{\text{outer}}(x)^2 - R_{\text{inner}}(x)^2 \right] dx
  \]
  Here, the outer curve is \( y = x \) and the inner curve is \( y = x^2 \).

  Therefore, the volume is:
  \[
  V = \pi \int_{0}^{1} \left[ (x)^2 - (x^2)^2 \right] dx
  \]
  Simplifying the integrand:
  \[
  V = \pi \int_{0}^{1} \left( x^2 - x^4 \right) dx
  \]

  Now, compute the integral:
  \[
  V = \pi \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_{0}^{1}
  \]
  Substituting the limits of integration:
  \[
  V = \pi \left( \frac{1^3}{3} - \frac{1^5}{5} \right) - \pi \left( 0 \right)
  \]
  \[
  V = \pi \left( \frac{1}{3} - \frac{1}{5} \right)
  \]
  \[
  V = \pi \left( \frac{5 - 3}{15} \right) = \frac{2\pi}{15}
  \]

Thus, the volume of the solid obtained by rotating the region around the **x-axis** is:
\[
V = \frac{2\pi}{15}
\]

---

### 2. Volume of the solid obtained by rotating the region about the **y-axis**:

For rotation around the y-axis, we use the **shell method** because we are rotating the region vertically.

- **Limits of integration**: As before, the region is bounded between \( x = 0 \) and \( x = 1 \).
  
- **Volume formula using shell method**: The volume \( V \) is given by:
  \[
  V = 2\pi \int_{a}^{b} \text{radius}(x) \cdot \text{height}(x) \, dx
  \]
  Here, the radius of each shell is \( x \), and the height of each shell is the difference between the curves \( y = x \) and \( y = x^2 \). So, the height is \( x - x^2 \).

  Therefore, the volume is:
  \[
  V = 2\pi \int_{0}^{1} x(x - x^2) \, dx
  \]
  Simplifying the integrand:
  \[
  V = 2\pi \int_{0}^{1} \left( x^2 - x^3 \right) dx
  \]

  Now, compute the integral:
  \[
  V = 2\pi \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_{0}^{1}
  \]
  Substituting the limits of integration:
  \[
  V = 2\pi \left( \frac{1^3}{3} - \frac{1^4}{4} \right) - 2\pi \left( 0 \right)
  \]
  \[
  V = 2\pi \left( \frac{1}{3} - \frac{1}{4} \right)
  \]
  \[
  V = 2\pi \left( \frac{4 - 3}{12} \right) = \frac{2\pi}{12} = \frac{\pi}{6}
  \]

Thus, the volume of the solid obtained by rotating the region around the **y-axis** is:
\[
V = \frac{\pi}{6}
\]

---

### Summary of results:
1. **Volume when rotating about the x-axis**: \( \frac{2\pi}{15} \)
2. **Volume when rotating about the y-axis**: \( \frac{\pi}{6} \)

---

Would you like a more detailed explanation or have any questions?

Here are some related questions for further exploration:
1. What happens if the region is rotated about the line \( y = 1 \)?
2. How would the volume change if the curves were shifted upwards by 1 unit?
3. How does the disk method compare to the washer method in these problems?
4. Can you find the centroid of the region enclosed by \( y = x \) and \( y = x^2 \)?
5. How does the shell method differ when rotating about a line other than the y-axis?

**Tip**: In problems involving rotations, carefully consider the method (disk/washer or shell) based on the axis of rotation for efficient problem-solving.

Learn how to find the volume of solids of revolution by rotating the region enclosed by y=x and y=x^2 around both the x-axis and y-axis. This problem covers the disk and shell methods, integral setup, and step-by-step solution. Suitable for students in Grades 11-12 and early college.

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