Find the volume of the solid generated by revolving the region bounded by the curves x = √y-1, y=3-x about the line y = 0

To find the volume of the solid generated by revolving the region bounded by the curves \( x = \sqrt{y-1} \) and \( y = 3 - x \) about the line \( y = 0 \), we'll use the method of disks (or washers).

### Step 1: Sketch the Region
- The curve \( x = \sqrt{y-1} \) can be rewritten as \( y = x^2 + 1 \) by squaring both sides and adding 1.
- The line \( y = 3 - x \) is a straight line with a negative slope.

### Step 2: Find the Intersection Points
To find the points of intersection, set the equations equal to each other:
\[
x^2 + 1 = 3 - x
\]
\[
x^2 + x - 2 = 0
\]
Factor the quadratic equation:
\[
(x + 2)(x - 1) = 0
\]
So, \( x = -2 \) or \( x = 1 \).

Now, find the corresponding \( y \)-values:
- For \( x = 1 \), \( y = 3 - 1 = 2 \).
- For \( x = -2 \), \( y = 3 - (-2) = 5 \).

Thus, the region is bounded between \( y = 2 \) and \( y = 5 \).

### Step 3: Set Up the Volume Integral
We use the method of washers, where the radius is determined by the distance from the curve to the axis of rotation (in this case, \( y = 0 \)).

- Outer radius \( R(y) = \) distance from \( y = 0 \) to the line \( x = \sqrt{y-1} \), so \( R(y) = y \).
- Inner radius \( r(y) = \) distance from \( y = 0 \) to the curve \( y = 3 - x \), which gives \( r(y) = 3 - y \).

The volume \( V \) is given by:
\[
V = \pi \int_{y=2}^{y=5} \left[ (\text{Outer Radius})^2 - (\text{Inner Radius})^2 \right] \, dy
\]

Substitute the expressions for the radii:
\[
V = \pi \int_{2}^{5} \left[ (\sqrt{y-1})^2 - (3 - y)^2 \right] dy
\]

### Step 4: Simplify and Integrate
\[
V = \pi \int_{2}^{5} \left[ (y-1) - (9 - 6y + y^2) \right] dy
\]
\[
V = \pi \int_{2}^{5} \left[ -y^2 + 7y - 10 \right] dy
\]

Integrate:
\[
V = \pi \left[ -\frac{y^3}{3} + \frac{7y^2}{2} - 10y \right]_{2}^{5}
\]

Evaluate at the limits \( y = 5 \) and \( y = 2 \):
\[
V = \pi \left[ \left(-\frac{125}{3} + \frac{175}{2} - 50\right) - \left(-\frac{8}{3} + \frac{28}{2} - 20\right) \right]
\]

Simplify the expression:
\[
V = \pi \left[-\frac{125}{3} + \frac{262.5}{3} - \frac{8}{3} + \frac{12}{3} \right]
\]

Final calculation gives:
\[
V = \frac{pi}{3} \left[ ( 24 ) - ( -24) \right] = \frac{48 \pi}{3}
\]

So the final answer for the volume is:

\[
V = \frac{48 \pi}{3} 
]

Would you like a detailed solution or need any more help with that?

Here are five related questions:
1. How would the problem change if the region was revolved around the line \( y = 1 \)?
2. What if the region was revolved around the line \( y = 4 \) instead?
3. How would you find the volume if the region was revolved around the \( y \)-axis?
4. What if you had to revolve the region around the line \( x = 3 \)?
5. How would you find the volume of a solid generated by revolving the same region around the line \( x = -1 \)?

Tip: Always visualize the region and the axis of rotation to better understand the setup of the integral.

Learn how to find the volume of the solid generated by revolving the region bounded by the curves x = √(y-1) and y = 3 - x about the line y = 0. Detailed step-by-step solution using the method of disks, including intersection points and integral setup. Suitable for advanced high school students studying integral calculus and volume of revolution.

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