To solve this problem, we need to compute the volume of the solid generated by rotating the region bounded by the given equations $y = -|x - 2| + 2$ and $y = 0$ about the $y$-axis.

Steps:

1. Understand the Shape of the Region:

   • $y = -|x - 2| + 2$ represents a "V"-shaped graph with its vertex at $(2, 2)$, symmetric about $x = 2$, decreasing linearly on both sides.

2. Boundaries of Integration:

   • Since $y = 0$ is the x-axis, the region is bounded between $y = 0$ and $y = -|x - 2| + 2$.
   • Solve for $x$ in terms of $y$: Split into two linear equations depending on $x$: $x = 2 - y \quad \text{(left branch)} \quad \text{and} \quad x = 2 + y \quad \text{(right branch)}.$

3. Volume via Shell Method:

   • The formula for the volume of a solid of revolution about the $y$-axis using the shell method is: $V = 2\pi \int_{y=0}^{y=2} (\text{radius}) \times (\text{height}) \, dy.$
   • The radius is $y$, and the height is $x_{\text{right}} - x_{\text{left}} = (2 + y) - (2 - y) = 2y$.

4. Set up the Integral: $V = 2\pi \int_{y=0}^{y=2} y \cdot 2y \, dy = 4\pi \int_{0}^{2} y^2 \, dy.$

5. Solve the Integral: $\int y^2 \, dy = \frac{y^3}{3}.$ Apply the limits $0$ to $2$: $\int_{0}^{2} y^2 \, dy = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}.$

6. Compute the Volume: $V = 4\pi \cdot \frac{8}{3} = \frac{32\pi}{3}.$

7. Numerical Approximation: Using $\pi \approx 3.14159$: $V \approx \frac{32 \cdot 3.14159}{3} \approx 33.51.$

Final Answer:

The volume of the solid is approximately 33.51 cubic units.

Let me know if you'd like further clarification or additional details. Here are some related questions:

1. How is the shell method different from the disk/washer method?
2. What happens if the solid is revolved about the $x$-axis instead?
3. Can this problem be solved using polar coordinates?
4. How does the absolute value in the function affect the region's boundaries?
5. What if the function was $y = -|x| + 2$ instead? Would the volume change?

Tip: Always sketch the region and axis of revolution to visualize the problem better before setting up the integral.