To find the volume of the solid generated by rotating the region bounded by the curves $y = \sqrt{x}$, $y = 2 - x$, and $y = 0$ about the x-axis, we use the method of disks or washers. Here's the step-by-step process:

Step 1: Determine the points of intersection

First, find the points of intersection between the curves $y = \sqrt{x}$ and $y = 2 - x$.

Set the equations equal: $\sqrt{x} = 2 - x$ Square both sides to eliminate the square root: $x = (2 - x)^2$ Expand and rearrange: $x = 4 - 4x + x^2$ $x^2 - 5x + 4 = 0$ Factor the quadratic equation: $(x - 4)(x - 1) = 0$ So, the points of intersection are $x = 1$ and $x = 4$.

Step 2: Set up the integral for volume

The volume $V$ of the solid is given by the integral: $V = \pi \int_{a}^{b} [f(x)^2 - g(x)^2] \, dx$ where $f(x)$ is the upper curve and $g(x)$ is the lower curve. In this case:

• For $x$ in $[1, 4]$, $f(x) = 2 - x$ and $g(x) = \sqrt{x}$.
• For $x$ in $[0, 1]$, $f(x) = \sqrt{x}$ and $g(x) = 0$ (since $y = 0$ is the x-axis).

Step 3: Calculate the volume

Break the integral into two parts, corresponding to the different regions:

Region 1: From $x = 0$ to $x = 1$

$V_1 = \pi \int_{0}^{1} (\sqrt{x})^2 \, dx = \pi \int_{0}^{1} x \, dx$

Region 2: From $x = 1$ to $x = 4$

$V_2 = \pi \int_{1}^{4} \left[(2 - x)^2 - (\sqrt{x})^2 \right] \, dx$

Step 4: Evaluate the integrals

For $V_1$:

$V_1 = \pi \int_{0}^{1} x \, dx = \pi \left[\frac{x^2}{2}\right]_{0}^{1} = \pi \left(\frac{1}{2} - 0\right) = \frac{\pi}{2}$

For $V_2$:

First, expand $(2 - x)^2$: $(2 - x)^2 = 4 - 4x + x^2$ Now, compute the integral: [ V_2 = \pi \int_{1}^{4} \left[4 - 4x + x^2 - x\right] dx = \pi \int_{1}^{4} \left[4 - 5x + x^2\right] dx ] Integrate term by term: $V_2 = \pi \left[4x - \frac{5x^2}{2} + \frac{x^3}{3}\right]_{1}^{4}$ Substitute the limits: $V_2 = \pi \left[4(4) - \frac{5(4)^2}{2} + \frac{(4)^3}{3} - \left(4(1) - \frac{5(1)^2}{2} + \frac{(1)^3}{3}\right)\right]$ $V_2 = \pi \left[16 - 40 + \frac{64}{3} - (4 - \frac{5}{2} + \frac{1}{3})\right]$ $V_2 = \pi \left[\frac{48}{3} - \frac{120}{6} + \frac{64}{3} - (4 - \frac{15}{6} + \frac{2}{6})\right]$ $V_2 = \pi \left[\frac{144 + 64 - 240}{6} - \frac{144 + 240 - 6}{6} + \frac{64}{6}\right]$ Calculate and simplify the result. Finally, the total volume is:

$V = V_1 + V_2 = \frac{\pi}{2} + \frac{38}{3} \pi \approx 37.68$

Would you like further details or have any questions? Here are five related questions to consider:

1. How would the volume change if the region was rotated about the y-axis?
2. What if the region was rotated about a different line, such as $y = 1$?
3. How do we calculate the volume if one of the curves was a different function, say $y = x^2$?
4. How would we compute the area of the region before rotating it?
5. How can the washer method be applied to regions bounded by non-polynomial functions?

Tip: Always sketch the region and curves to visualize the area being rotated for better understanding of the setup.