To find the volume of the solid generated by revolving the region bounded by the curves $y = \sqrt{x}$, $y = 0$, and $y = x - 2$ about the x-axis using the shell method, we can follow these steps:

Step 1: Identify the Region of Interest

First, we need to find the intersection points of the curves $y = \sqrt{x}$ and $y = x - 2$ since they will define the limits of integration.

Set $\sqrt{x} = x - 2$ and solve for $x$: $\sqrt{x} = x - 2$ Square both sides: $x = (x - 2)^2$ Expand and simplify: $x = x^2 - 4x + 4$ $0 = x^2 - 5x + 4$ $x^2 - 5x + 4 = 0$ This quadratic equation can be solved using the quadratic formula: $x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(4)}}{2(1)}$ $x = \frac{5 \pm \sqrt{25 - 16}}{2}$ $x = \frac{5 \pm \sqrt{9}}{2}$ $x = \frac{5 \pm 3}{2}$ $x = 4 \quad \text{or} \quad x = 1$

So, the intersection points are at $x = 1$ and $x = 4$.

Step 2: Set Up the Integral for the Volume Using the Shell Method

For the shell method, the formula for the volume when revolving around the x-axis is: $V = 2\pi \int_{a}^{b} (y \cdot f(y)) \, dy$ where $y$ is the radius and $f(y)$ is the height of the shell.

For the given curves:

• The region is bounded between $y = 0$ and $y = 2$ (since $y = \sqrt{4} = 2$).
• The height of the shell is given by the difference $f(y) = (y + 2) - y^2$.

Thus, the volume integral is: $V = 2\pi \int_{0}^{2} y \cdot \left[(y + 2) - y^2 \right] dy$ Simplify the integrand: $V = 2\pi \int_{0}^{2} \left(y^2 + 2y - y^3\right) dy$

Step 3: Evaluate the Integral

Now, we integrate: $V = 2\pi \left[\frac{y^3}{3} + y^2 - \frac{y^4}{4} \right]_{0}^{2}$

Substitute the limits: $V = 2\pi \left[\left(\frac{2^3}{3} + 2^2 - \frac{2^4}{4}\right) - \left(\frac{0^3}{3} + 0^2 - \frac{0^4}{4}\right)\right]$ $V = 2\pi \left[\left(\frac{8}{3} + 4 - 4\right) - 0\right]$ $V = 2\pi \cdot \frac{8}{3}$ $V = \frac{16\pi}{3}$

Final Answer

The volume of the solid generated by revolving the region bounded by the curves about the x-axis is $\frac{16\pi}{3}$ cubic units.

Would you like more details or have any questions?

Here are 5 related questions to consider:

1. How does the shell method differ from the disk/washer method in terms of setup?
2. Can the shell method be used for solids of revolution around lines other than the x-axis?
3. How would the problem change if the region were revolved around the y-axis instead?
4. What are the benefits of using the shell method in certain cases over the disk/washer method?
5. What would be the setup for the volume if the region was bounded by $y = x$ instead of $y = \sqrt{x}$?

Tip: When using the shell method, the radius of the shell is the distance from the axis of revolution to the strip, and the height of the shell is the length of the strip perpendicular to the axis.