To determine the volume of this solid, we can use the formula for the volume of a solid with a square cross-section. Here are the steps:

1. The cross-sectional area of the square perpendicular to the x-axis at a point $x$ will be the square of the distance between the two curves $y = x$ and $y = 4$, as this distance forms the base of the square.

2. The distance between $y = x$ and $y = 4$ is: $\text{Base length} = 4 - x$

3. Since the cross-section is a square, the area of the cross-section at any point $x$ is: $A(x) = (4 - x)^2$

4. To find the volume, we integrate this area over the interval $x \in [-2, 2]$: $V = \int_{-2}^{2} A(x) \, dx = \int_{-2}^{2} (4 - x)^2 \, dx$

5. Expanding the integrand: $(4 - x)^2 = 16 - 8x + x^2$

6. Substitute into the integral: $V = \int_{-2}^{2} (16 - 8x + x^2) \, dx$

7. Now, we can evaluate each term separately: $V = \int_{-2}^{2} 16 \, dx - \int_{-2}^{2} 8x \, dx + \int_{-2}^{2} x^2 \, dx$

   • For $\int_{-2}^{2} 16 \, dx = 16 \cdot (2 - (-2)) = 16 \cdot 4 = 64$
   • For $\int_{-2}^{2} 8x \, dx$, this is zero because it’s an odd function over a symmetric interval.
   • For $\int_{-2}^{2} x^2 \, dx = \frac{2}{3} x^3 \Big|_{-2}^{2} = \frac{2}{3} (8) = \frac{16}{3}$

8. Adding these up: $V = 64 + 0 + \frac{16}{3} = \frac{192 + 16}{3} = \frac{208}{3} = 64$

So the correct answer is:

c. 64

Would you like a more detailed breakdown of any step?

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Here are some additional questions that extend this problem:

1. How would the volume change if the base of the square was between $y = x$ and $y = 3$?
2. How does changing the interval affect the volume of the solid?
3. What if the cross-sectional shape were a rectangle instead of a square?
4. How would you find the volume if the cross-sections were semicircles instead of squares?
5. How does the method of finding volume change if we rotate the region around the x-axis?

Tip: When calculating volumes with cross-sectional areas, ensure to correctly express the distance or function defining the shape based on the given constraints.