Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves y=x^2,0<=x<=5,y=25, and x=0

To find the volume of the solid formed by rotating the region bounded by \( y = x^2 \), \( y = 25 \), \( x = 0 \), and \( x = 5 \) around the \( y \)-axis, we can use the disk or washer method.

### Step-by-Step Solution Using the Washer Method:

1. **Set up the boundaries and the functions:**
   - We have \( y = x^2 \), which describes the curve, and \( y = 25 \), which is a horizontal line.
   - The region is bounded vertically from \( y = 0 \) to \( y = 25 \).

2. **Solve for \( x \) in terms of \( y \):**
   - Since \( y = x^2 \), we get \( x = \sqrt{y} \) (only the positive root since \( x \geq 0 \)).

3. **Volume of Rotation Using the Washer Method:**
   - The washers have an inner radius of \( r_{\text{inner}} = 0 \) and an outer radius of \( r_{\text{outer}} = \sqrt{y} \).
   - The volume \( V \) of the solid is given by integrating the area of each washer as it rotates around the \( y \)-axis from \( y = 0 \) to \( y = 25 \).

4. **Set up the integral:**
   \[
   V = \int_{0}^{25} \pi \left( \text{outer radius}^2 - \text{inner radius}^2 \right) \, dy
   \]
   Substituting the values for the radii, we get:
   \[
   V = \int_{0}^{25} \pi \left( (\sqrt{y})^2 - 0^2 \right) \, dy
   \]
   Simplifying inside the integral:
   \[
   V = \int_{0}^{25} \pi y \, dy
   \]

5. **Integrate:**
   - Factor out \( \pi \):
   \[
   V = \pi \int_{0}^{25} y \, dy
   \]
   - Compute the integral:
   \[
   V = \pi \left[ \frac{y^2}{2} \right]_{0}^{25}
   \]
   - Substitute the limits:
   \[
   V = \pi \left( \frac{25^2}{2} - \frac{0^2}{2} \right)
   \]
   - Calculate \( 25^2 = 625 \):
   \[
   V = \pi \cdot \frac{625}{2}
   \]
   - Simplify:
   \[
   V = \frac{625\pi}{2}
   \]

### Final Answer:
\[
V = \frac{625\pi}{2} \text{ cubic units}
\]

Would you like more details, or do you have any questions about the steps?

### Related Questions
1. How would the process change if we rotated the region around the \( x \)-axis?
2. What if the upper boundary were \( y = 16 \) instead of \( y = 25 \)?
3. How does the choice of radius (inner or outer) affect the integral setup in washer problems?
4. Could this problem be solved using cylindrical shells? If so, how?
5. What happens if we rotate this region about the line \( x = 6 \) instead of the \( y \)-axis?

### Tip:
For problems involving rotation around an axis, clearly distinguish between the inner and outer radius for the washer method to avoid calculation errors.

Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves y=x^2, 0<=x<=5, y=25, and x=0

Learn how to calculate the volume of a solid obtained by rotating the region bounded by y=x^2, y=25, 0≤x≤5 around the y-axis using the washer method. This solution includes step-by-step calculations and integration, ideal for Calculus II students.

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