The given solid has a circular disk with radius $r$ as its base, and parallel cross-sections perpendicular to the base are squares. We are tasked with finding the volume of this solid by setting up and evaluating an integral.

Step-by-step explanation:

1. Base of the Solid: The base is a circular disk of radius $r$. The equation of the circle (viewed from the side along the x-axis) can be described as: $x^2 + y^2 = r^2$ Solving for $y$, we have: $y = \sqrt{r^2 - x^2}$ This describes the upper half of the circular base (since $y \geq 0$).

2. Cross-sectional Area: Each cross-section perpendicular to the x-axis is a square. The side length of the square at any point $x$ is $2y$, where $y = \sqrt{r^2 - x^2}$. So, the side length of the square is: $\text{Side length} = 2\sqrt{r^2 - x^2}$ Therefore, the area $A(x)$ of the square at a point $x$ is: $A(x) = \left(2\sqrt{r^2 - x^2}\right)^2 = 4(r^2 - x^2)$

3. Volume of the Solid: To find the total volume of the solid, we integrate the area of the cross-sections from $x = -r$ to $x = r$: $V = \int_{-r}^{r} A(x) \, dx = \int_{-r}^{r} 4(r^2 - x^2) \, dx$

4. Simplifying the Integral: We can simplify this integral as: $V = 4 \int_{-r}^{r} (r^2 - x^2) \, dx$ Since the function $r^2 - x^2$ is symmetric, we can double the integral from $0$ to $r$: $V = 8 \int_{0}^{r} (r^2 - x^2) \, dx$

5. Solving the Integral: Now, we solve the integral: $\int (r^2 - x^2) \, dx = r^2x - \frac{x^3}{3}$ Evaluating this from $0$ to $r$: $V = 8 \left[ r^2x - \frac{x^3}{3} \right]_0^r = 8 \left( r^3 - \frac{r^3}{3} \right)$ Simplifying: $V = 8 \left( \frac{3r^3}{3} - \frac{r^3}{3} \right) = 8 \times \frac{2r^3}{3} = \frac{16r^3}{3}$

Final Answer:

The volume of the solid is: $V = \frac{16r^3}{3}$

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Would you like a more detailed explanation, or do you have any questions?

Here are 5 related questions to deepen your understanding:

1. How would the volume formula change if the cross-sections were equilateral triangles instead of squares?
2. What if the base was an ellipse rather than a circle? How would you set up the volume integral?
3. Can you compute the surface area of the solid given this setup?
4. How would you handle this problem if the cross-sections were semicircles instead of squares?
5. What happens to the volume if you double the radius of the base?

Tip: When dealing with volumes of solids with varying cross-sections, always start by identifying the area of the cross-section and integrating along the axis perpendicular to it.