The problem describes a solid whose base is the region bounded by the ellipse $x^2 + 4y^2 = 4$, and all vertical cross-sections parallel to the $x$-axis are squares. We are tasked with finding the volume of the solid.

Step 1: Analyze the Base

The equation of the ellipse is given by:

$x^2 + 4y^2 = 4$

To rewrite this in a more useful form, divide through by 4:

$\frac{x^2}{4} + \frac{y^2}{1} = 1$

This is an ellipse centered at the origin with semi-major axis 2 along the $x$-axis and semi-minor axis 1 along the $y$-axis.

Step 2: Set Up the Cross-Sectional Area

Since all vertical cross-sections parallel to the $x$-axis are squares, the side length of each square is determined by the distance between the top and bottom of the ellipse at any given $x$-value.

For a given $x$, we solve for $y$ in the equation of the ellipse:

$x^2 + 4y^2 = 4$ $4y^2 = 4 - x^2$ $y^2 = \frac{4 - x^2}{4}$ $y = \pm \sqrt{\frac{4 - x^2}{4}} = \pm \frac{\sqrt{4 - x^2}}{2}$

The distance between the top and bottom of the ellipse at a given $x$ is:

$2y = 2 \times \frac{\sqrt{4 - x^2}}{2} = \sqrt{4 - x^2}$

Thus, the side length of the square cross-section at $x$ is $s(x) = \sqrt{4 - x^2}$.

Step 3: Volume of the Solid

The volume of the solid is found by integrating the area of the square cross-section along the $x$-axis. The area of a square with side length $s(x)$ is:

$A(x) = s(x)^2 = (\sqrt{4 - x^2})^2 = 4 - x^2$

The volume $V$ is the integral of the area function from $x = -2$ to $x = 2$ (since the ellipse has semi-major axis 2 along the $x$-axis):

$V = \int_{-2}^{2} (4 - x^2) \, dx$

Step 4: Perform the Integration

Let's compute the integral:

$V = \int_{-2}^{2} 4 \, dx - \int_{-2}^{2} x^2 \, dx$

For the first integral:

$\int_{-2}^{2} 4 \, dx = 4 \times (2 - (-2)) = 4 \times 4 = 16$

For the second integral, since $x^2$ is an even function:

$\int_{-2}^{2} x^2 \, dx = 2 \int_{0}^{2} x^2 \, dx = 2 \times \left[ \frac{x^3}{3} \right]_{0}^{2} = 2 \times \left( \frac{8}{3} - 0 \right) = \frac{16}{3}$

Thus, the volume is:

$V = 16 - \frac{16}{3} = \frac{48}{3} - \frac{16}{3} = \frac{32}{3}$

Final Answer:

The volume of the solid is $\boxed{\frac{32}{3}}$.

Would you like to explore any details further, or do you have any questions on the process? Here are some related questions you can consider:

1. How does the shape of the base affect the volume calculation?
2. What would the volume be if the cross-sections were triangles instead of squares?
3. How do you compute the area of other common cross-sectional shapes (like circles or rectangles)?
4. How would the volume change if the ellipse had different dimensions (e.g., semi-major axis 3 and semi-minor axis 2)?
5. How do you handle solids with curved boundaries in general, and are there shortcuts for certain shapes?

Tip: When dealing with solids of known cross-section, always focus on finding the area function of the cross-section first, and then integrate over the region.