To find the volume of the solid generated by revolving the region bounded by the curve $y = \sqrt{16 - x^2}$ and the line $y = 0$ about the x-axis, we can use the disk method.

1. Identify the Region to Revolve: The curve $y = \sqrt{16 - x^2}$ represents the upper half of a circle with radius 4, centered at the origin. This curve intersects the x-axis (where $y = 0$) at $x = -4$ and $x = 4$.

2. Set Up the Integral for the Volume: The volume $V$ of the solid formed by revolving this region about the x-axis is given by: $V = \pi \int_{-4}^{4} \left( \sqrt{16 - x^2} \right)^2 \, dx$

3. Simplify the Integral: Since $\left( \sqrt{16 - x^2} \right)^2 = 16 - x^2$, the volume integral becomes: $V = \pi \int_{-4}^{4} (16 - x^2) \, dx$

4. Evaluate the Integral: Split the integral: $V = \pi \left( \int_{-4}^{4} 16 \, dx - \int_{-4}^{4} x^2 \, dx \right)$

5. Calculate Each Part:

   • For $\int_{-4}^{4} 16 \, dx$: $\int_{-4}^{4} 16 \, dx = 16 \cdot (4 - (-4)) = 16 \cdot 8 = 128$
   • For $\int_{-4}^{4} x^2 \, dx$: Because $x^2$ is an even function, we can compute this integral from $0$ to $4$ and double it: $\int_{-4}^{4} x^2 \, dx = 2 \int_{0}^{4} x^2 \, dx = 2 \cdot \left[ \frac{x^3}{3} \right]_{0}^{4} = 2 \cdot \frac{64}{3} = \frac{128}{3}$

6. Combine Results: Substitute back into the volume formula: $V = \pi \left( 128 - \frac{128}{3} \right) = \pi \cdot \frac{384 - 128}{3} = \pi \cdot \frac{256}{3} = \frac{256\pi}{3}$

Final Answer:

The volume of the solid is: $V = \frac{256\pi}{3}$

Would you like further clarification on any part of this process?

Related Questions:

1. How would the solution change if the region were revolved around the y-axis instead?
2. What is the volume if we revolve only the first quadrant portion of the region?
3. How do we apply the washer method for a similar volume problem?
4. Can this integral be solved using polar coordinates, and what would that look like?
5. What would the setup be if we revolved this region about a line $y = c$ where $c > 0$?

Tip: For regions that are symmetric about an axis, consider integrating only half the region and then doubling the result to simplify calculations.