Problem 12 - Thomas Calculus

Question

Cavalieri's Principle A solid lies between planes perpendicular to the \(x\) -axis at \(x=0\) and \(x=12 .\) The cross-sections by planes perpendicular to the \(x\) -axis are circular disks whose diameters run from the line \(y=x / 2\) to the line \(y=x\) as shown in the accompanying figure. Explain why the solid has the same volume as a right circular cone with base radius 3 and height \(12 .\)

Step-by-Step Solution

Answer

The solid and the cone have the same volume by Cavalieri's Principle, both being \(36\pi\).

Step 1: Identifying the cross-section shape

The solid described in the problem has cross-sections that are circular disks. Each disk's diameter spans from the line \(y = \frac{x}{2}\) to \(y = x\).

Step 2: Find the radius of a cross-section

The diameter of each cross-section disk at a particular value of \(x\) is given by \(x - \frac{x}{2} = \frac{x}{2}\). Thus, the radius \(r\) of each circular disk is half of the diameter: \(r = \frac{x}{4}\).

Step 3: Set up the volume integral

The volume \(V\) of the solid can be calculated by integrating the area of the circular disks along the \(x\)-axis from \(x = 0\) to \(x = 12\). The area \(A(x)\) of each circular cross-section is \(\pi r^2 = \pi \left(\frac{x}{4}\right)^2\).

Step 4: Calculate the volume integral

Integrate the area function from \(0\) to \(12\):\[V = \int_{0}^{12} \pi \left(\frac{x}{4}\right)^2 \, dx = \pi \int_{0}^{12} \frac{x^2}{16} \, dx = \frac{\pi}{16} \int_{0}^{12} x^2 \, dx.\]

Step 5: Solve the integral

Calculate the integral:\[\frac{\pi}{16} \left[ \frac{x^3}{3} \right]_{0}^{12} = \frac{\pi}{16} \cdot \frac{12^3}{3} = \frac{\pi}{16} \cdot \frac{1728}{3} = \frac{\pi}{16} \cdot 576 = 36\pi.\]

Step 6: Comparing volumes with the cone

The volume of the right circular cone with base radius \(3\) and height \(12\) is given by \(\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (3^2)(12) = 36\pi\). The calculated solid volume \(36\pi\) matches this cone volume, proving the volumes are the same.

Key Concepts

Volume Integral

In the context of this exercise, a volume integral is a powerful tool used to figure out the volume of a solid by integrating the area of its cross-sections along an axis. In simpler terms, it involves summing up the tiny volumes created by slicing the solid into thin pieces and considering each piece's surface area. Reflecting on the original problem, the volume integral was critical because it allowed us to calculate the volume of the solid with non-constant radii across its length.

The integration limits are given by the position where the solid starts and ends. Here, it's from 0 to 12 along the x-axis. Using the formula for the area of a circle, \(A(x) = \pi r^{2}\), with \(r = \frac{x}{4}\), we can calculate each cross-section's area and integrate these areas to find the total volume of the solid.

Cross-section

Cross-sections are slices of the solid taken perpendicular to a particular axis. You can think of them as looking inside a loaf of bread where each slice represents a cross-section. In this exercise, cross-sections are taken perpendicular to the x-axis, meaning they reflect the shape of the solid at different points along that axis.

The problem specifies that each cross-section is a circular disk, with its diameter stretching between two specified lines: \(y = \frac{x}{2}\) and \(y = x\). As you move along the x-axis, this diameter changes. By finding the area of these slices, using the formula for the area of a circle, we can integrate this area over the given range (in this case from x = 0 to x = 12) to calculate the solid's volume.

Circular Disks

The concept of circular disks is central here as it forms the foundation for calculating the volume of the solid. A circular disk is essentially a flat, round piece with a certain radius. In the provided exercise, the solid's cross-sections are these circular disks.

Each disk's diameter is determined by the distance from the line \(y = \frac{x}{2}\) to the line \(y = x\). The difference \(x - \frac{x}{2} = \frac{x}{2}\) gives the diameter of each disk. To find the radius r, simply halve this diameter: \(r = \frac{x}{4}\). This radius varies with each x value, mirroring how the solid tapers as it extends along the x-axis. With this radius, the area of each disk can be determined and later summed, through integration, to get the solid's volume.

Right Circular Cone

A right circular cone is a three-dimensional geometrical shape with a circular base and a single vertex (the apex) where all straight lines drawn from the base to the apex form right angles with the base. This specific geometry makes it easier to determine the volume of a cone via a formula.

In the solution, the volume of the target solid (formed by stacks of circular disks) is found equal to that of a right circular cone with a base radius of 3 and a height of 12. This is confirmed through computation. The volume of a cone is given by the formula: \(\frac{1}{3}\pi r^{2} h\) where \(r\) is the base radius, and \(h\) is the height. By plugging in the values \(r = 3\) and \(h = 12\), we find the cone's volume as \(36\pi\), identical to the computed volume of the solid, thereby verifying Cavalieri's Principle in this case.