The image describes a solid with a circular disk base and parallel cross-sections perpendicular to the base that are squares. The problem asks to:

1. Set up an integral to find the volume $V$ of the solid.
2. Calculate the volume.

Step 1: Understanding the Geometry

• The base of the solid is a circular disk with radius $r$.
• The cross-sections are squares perpendicular to the base.

Setting Up the Integral

The area of a square depends on the length of its side, which will vary based on its position on the $x$-axis. For a circular base, the equation of the circle is: $x^2 + y^2 = r^2$ For any cross-section, the side of the square is twice the value of $y$, which is the distance from the center of the base to the edge of the circle. So, $y = \sqrt{r^2 - x^2}$, and the side of the square is: $\text{Side length of square} = 2y = 2\sqrt{r^2 - x^2}$

The area of a square cross-section is: $\text{Area} = \left(2\sqrt{r^2 - x^2}\right)^2 = 4(r^2 - x^2)$

Volume Integral

To find the volume, we integrate the area of the cross-sections along the $x$-axis from $-r$ to $r$: $V = \int_{-r}^{r} 4(r^2 - x^2) \, dx$

Step 2: Solving the Integral

We simplify and solve the integral: $V = 4 \int_{-r}^{r} (r^2 - x^2) \, dx$ This can be split into two parts: $V = 4 \left[ \int_{-r}^{r} r^2 \, dx - \int_{-r}^{r} x^2 \, dx \right]$

First Integral:

$\int_{-r}^{r} r^2 \, dx = r^2 \int_{-r}^{r} dx = r^2 [x]_{-r}^{r} = r^2(2r) = 2r^3$

Second Integral:

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

Now, substitute both integrals: $V = 4 \left( 2r^3 - \frac{2r^3}{3} \right) = 4 \left( \frac{6r^3}{3} - \frac{2r^3}{3} \right) = 4 \times \frac{4r^3}{3} = \frac{16r^3}{3}$

Final Answer:

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

Would you like more details on any part of this process?

Related Questions:

1. How is the area of the square related to the equation of the circle?
2. What changes if the cross-sections are triangles instead of squares?
3. Can you set up a similar volume integral for a different 3D solid, like a cone?
4. How does symmetry help in solving these types of problems?
5. What are the limits of integration for a solid with a non-circular base?

Tip:

When setting up volume integrals, always relate the geometry of the cross-section to the equation of the base!