Find the volume of the solid obtained by revolving the region bounded by y=5/8x-5/8x^2 and the x-axis around the x-axis

To find the volume of the solid obtained by revolving the region bounded by the curve \( y = \frac{5}{8}x - \frac{5}{8}x^2 \) and the x-axis around the x-axis, we can use the **Disk Method**. 

### Steps:
1. **Identify the bounds**:
   The curve is given by \( y = \frac{5}{8}x - \frac{5}{8}x^2 \), and it's bounded by the x-axis, so we need to find the intersection points between the curve and the x-axis (i.e., where \( y = 0 \)).
   
   Solve \( \frac{5}{8}x - \frac{5}{8}x^2 = 0 \):
   \[
   \frac{5}{8}x(1 - x) = 0
   \]
   This gives \( x = 0 \) and \( x = 1 \) as the bounds.

2. **Set up the volume integral**:
   The formula for the volume using the Disk Method is:
   \[
   V = \pi \int_{a}^{b} [f(x)]^2 \, dx
   \]
   where \( f(x) = \frac{5}{8}x - \frac{5}{8}x^2 \) and the limits of integration are from \( x = 0 \) to \( x = 1 \).

   Thus, the volume integral becomes:
   \[
   V = \pi \int_{0}^{1} \left( \frac{5}{8}x - \frac{5}{8}x^2 \right)^2 \, dx
   \]

3. **Expand the integrand**:
   First, square the expression \( \frac{5}{8}x - \frac{5}{8}x^2 \):
   \[
   \left( \frac{5}{8}x - \frac{5}{8}x^2 \right)^2 = \left( \frac{5}{8} \right)^2 \left( x - x^2 \right)^2 = \frac{25}{64} (x^2 - 2x^3 + x^4)
   \]

4. **Integrate term-by-term**:
   Now, integrate each term from 0 to 1:
   \[
   \int_{0}^{1} x^2 \, dx = \frac{1}{3}, \quad \int_{0}^{1} x^3 \, dx = \frac{1}{4}, \quad \int_{0}^{1} x^4 \, dx = \frac{1}{5}
   \]

   Substituting these into the expression:
   \[
   V = \pi \times \frac{25}{64} \left( \frac{1}{3} - 2 \times \frac{1}{4} + \frac{1}{5} \right)
   \]

5. **Simplify the expression**:
   Simplify the terms inside the parentheses:
   \[
   \frac{1}{3} - 2 \times \frac{1}{4} + \frac{1}{5} = \frac{1}{3} - \frac{1}{2} + \frac{1}{5}
   \]
   Find a common denominator (30):
   \[
   \frac{1}{3} = \frac{10}{30}, \quad \frac{1}{2} = \frac{15}{30}, \quad \frac{1}{5} = \frac{6}{30}
   \]
   So:
   \[
   \frac{10}{30} - \frac{15}{30} + \frac{6}{30} = \frac{1}{30}
   \]

6. **Final expression for volume**:
   Now substitute this back:
   \[
   V = \pi \times \frac{25}{64} \times \frac{1}{30} = \frac{25\pi}{1920}
   \]

### Final Answer:
The volume of the solid is:
\[
V = \frac{25\pi}{1920} \, \text{cubic units}
\]

Do you want more details or have any questions?

### Related Questions:
1. How is the disk method different from the washer method in volume calculation?
2. What if the curve \( y = \frac{5}{8}x - \frac{5}{8}x^2 \) was revolved around the y-axis instead of the x-axis?
3. Can we calculate the surface area of the solid formed by revolution using a different method?
4. How does the presence of an upper boundary change the limits in volume calculations?
5. What happens if the curve is revolved around a different line, such as \( y = 1 \) or \( x = 2 \)?

### Tip:
When using the disk method, always verify that the curve intersects the axis of revolution at the limits of integration, as this simplifies the setup significantly.

Find the volume of the solid obtained by revolving the region bounded by y = 5/8x - 5/8x^2 and the x-axis around the x-axis

Find the volume of the solid obtained by revolving the region bounded by the curve y = 5/8x - 5/8x^2 and the x-axis around the x-axis. This problem uses the disk method and covers key concepts in calculus, including integration and volume calculation. Suitable for Grade 12 and college-level calculus students.

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