To solve this problem using the disc or washer method, we'll consider the volume generated by revolving the region around the y-axis. In this case, we'll use the washer method because we are dealing with a region bounded by an outer and inner function with respect to the axis of rotation.

### Step 1: Express \( x \) as a function of \( y \)

The given equations are:
- \( y = 25 - x^2 \)
- \( y = 0 \) (the x-axis)
- \( x = 2 \) and \( x = 5 \) (vertical lines)

First, solve for \( x \) from the equation \( y = 25 - x^2 \):

\[
x^2 = 25 - y \implies x = \sqrt{25 - y}
\]

Since we are revolving around the y-axis, we will express everything in terms of \( y \).

### Step 2: Determine the limits for \( y \)

The region is bounded vertically between the points where \( x = 2 \) and \( x = 5 \). Calculate the corresponding \( y \) values:

For \( x = 2 \):

\[
y = 25 - 2^2 = 25 - 4 = 21
\]

For \( x = 5 \):

\[
y = 25 - 5^2 = 25 - 25 = 0
\]

So, the limits for \( y \) will be from 0 to 21.

### Step 3: Set up the volume integral

Using the washer method, the volume \( V \) is given by:

\[
V = \pi \int_{y_1}^{y_2} \left[ (R_{outer})^2 - (R_{inner})^2 \right] \, dy
\]

Where:
- \( R_{outer} \) is the radius of the outer function (for the vertical lines \( x = 5 \), which is constant).
- \( R_{inner} \) is the radius of the inner function, which corresponds to the curve \( x = \sqrt{25 - y} \).

For this problem:
- \( R_{outer} = 5 \) (the distance from \( x = 5 \) to the y-axis).
- \( R_{inner} = 2 \) (the distance from \( x = 2 \) to the y-axis).

Thus, the integral becomes:

\[
V = \pi \int_{0}^{21} \left[ (5)^2 - (2)^2 \right] \, dy
\]

Simplify the expression inside the integral:

\[
V = \pi \int_{0}^{21} \left[ 25 - 4 \right] \, dy = \pi \int_{0}^{21} 21 \, dy
\]

### Step 4: Integrate

Now, integrate with respect to \( y \):

\[
V = \pi \times 21 \times \left[ y \right]_{0}^{21} = \pi \times 21 \times (21 - 0) = \pi \times 21 \times 21 = 441\pi
\]

### Final Answer:
The volume of the solid generated by revolving the region about the y-axis is \( \boxed{441\pi} \) cubic units.

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

Here are 5 related questions to explore further:
1. What would be the volume if the region were revolved around the x-axis instead of the y-axis?
2. How does the method of washers compare to the method of cylindrical shells for this problem?
3. How would the volume change if the region were bounded by \( x = 1 \) instead of \( x = 2 \)?
4. What is the effect on the volume if the parabola was \( y = 36 - x^2 \) instead?
5. Can you compute the surface area of the solid formed by this revolution?

**Tip:** The washer method is most effective when dealing with regions that are not directly adjacent to the axis of rotation. Be sure to correctly identify the outer and inner radii.

Learn how to calculate the volume using the disc or washer method in integral calculus. Detailed step-by-step solution provided for the problem involving revolving a region around the y-axis. Suitable for college-level students studying integral calculus.

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