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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?

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**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.

Find the volume of the solid generated by revolving the region bounded by the curve y = √(16 - x^2) and y = 0 about the x-axis.

Calculate the volume of the solid generated by revolving the region bounded by y = √(16 - x^2) and y = 0 about the x-axis. This problem explores the disk method, integration techniques, and volume of solids of revolution, suitable for Grade 12 and college calculus students.

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