Find the volume of the solid obtained by rotating the region bounded by the curves x=y−y2
 and x=0
 about the y
-axis.

To find the volume of the solid obtained by rotating the region bounded by the curves \( x = y - y^2 \) and \( x = 0 \) about the \( y \)-axis, we can use the method of disks or washers.

### Step 1: Understand the problem

The region is bounded by:
- \( x = y - y^2 \) (a parabola)
- \( x = 0 \) (the y-axis)

We are rotating this region about the \( y \)-axis.

### Step 2: Set up the volume integral

The volume \( V \) of the solid formed by rotating this region about the \( y \)-axis can be computed using the formula:

\[
V = \pi \int_{c}^{d} \left[ R(y) \right]^2 \, dy
\]

Where \( R(y) \) is the distance from the \( y \)-axis to the curve (i.e., \( R(y) = x = y - y^2 \)).

### Step 3: Determine the bounds of integration

The curve \( x = y - y^2 \) intersects the \( y \)-axis where \( x = 0 \):

\[
y - y^2 = 0
\]

Factoring, we get:

\[
y(y - 1) = 0
\]

So, \( y = 0 \) and \( y = 1 \). These are the limits of integration \( c \) and \( d \), so the integration will be from \( y = 0 \) to \( y = 1 \).

### Step 4: Compute the volume integral

Substitute \( R(y) = y - y^2 \) into the volume formula:

\[
V = \pi \int_{0}^{1} \left( y - y^2 \right)^2 \, dy
\]

Now, expand the square:

\[
V = \pi \int_{0}^{1} \left( y^2 - 2y^3 + y^4 \right) \, dy
\]

### Step 5: Integrate

Integrate each term:

\[
V = \pi \left[ \frac{y^3}{3} - \frac{2y^4}{4} + \frac{y^5}{5} \right]_{0}^{1}
\]

Simplifying the expression:

\[
V = \pi \left[ \frac{1}{3} - \frac{1}{2} + \frac{1}{5} \right]
\]

\[
V = \pi \left( \frac{10}{30} - \frac{15}{30} + \frac{6}{30} \right)
\]

\[
V = \pi \left( \frac{1}{30} \right)
\]

\[
V = \frac{\pi}{30}
\]

### Final Answer

The volume of the solid is \( \boxed{\frac{\pi}{30}} \).

Would you like more details on any step, or do you have any questions?

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Here are 5 related questions:
1. How would the volume change if the region were rotated about the x-axis instead of the y-axis?
2. What if the curve were \( x = y^2 - y \) instead of \( x = y - y^2 \)?
3. How would you find the surface area of the solid obtained by rotating the same region about the y-axis?
4. How can the disk method be applied if the region is bounded by different curves?
5. How does the method change if the solid is formed by rotating around a line other than the y-axis?

**Tip:** Always ensure that you carefully determine the bounds of integration, as they can significantly impact your final result.

Learn how to find the volume of a solid obtained by rotating the region bounded by x=y−y^2 and x=0 about the y-axis. Detailed step-by-step solution using calculus and integration. Suitable for advanced high school and college students.

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