Problem 38 - Calculus and its applications

Question

Using volume by shells, prove that the volume of a right circular cone of height \(h\) and radius \(r\) is \(V=\frac{1}{3} \pi r^{2} h\).

Step-by-Step Solution

Answer

Using shell method, the cone's volume is \(V=\frac{1}{3}\pi r^2 h\).

Step 1: Understand the Problem

We need to derive the formula for the volume of a right circular cone using the method of cylindrical shells. The cone has a height \(h\) and a base radius \(r\).

Step 2: Set Up the Integral for Volume Using Shells

Using the method of cylindrical shells, we generate cylindrical shells by revolving vertical slices around the y-axis. Each shell has a height, a radius, and a thickness. The radius is \(x\), the height is given by the equation of line from the base to the apex \(y=\frac{h}{r}x\), and thickness as \(dx\). The circumference of the shell is \(2\pi x\).

Step 3: Write the Expression for the Volume of a Single Shell

The volume of a thin shell is given by \(dV = 2\pi x \cdot \text{height} \cdot \text{thickness} = 2\pi x \cdot \frac{h}{r}x \cdot dx\). This simplifies to \(dV = \frac{2\pi hx^2}{r} dx\).

Step 4: Set Limits for Integration

The shells run from \(x=0\) to \(x=r\) (the base extends from the center to the edge of the cone). Thus, the volume \(V\) is the integral from 0 to \(r\).

Step 5: Evaluate the Integral

Integrate \(\frac{2\pi hx^2}{r}\) from 0 to \(r\): \[V = \frac{2\pi h}{r} \int_0^r x^2 \, dx\]Calculate the integral: \[\int_0^r x^2 \, dx = \left. \frac{x^3}{3} \right|_0^r = \frac{r^3}{3}\]Insert back into the volume integral: \[V = \frac{2\pi h}{r} \cdot \frac{r^3}{3} = \frac{2\pi hr^2}{3}\]

Step 6: Simplify the Result

Finally, the above result is \(V = \frac{1}{3}\pi r^2 h\) confirming the standard formula for the volume of a cone.

Key Concepts

Calculus

Calculus is a branch of mathematics focused on change and motion, essential when dealing with dynamic quantities. In this problem, calculus helps determine the volume of a right circular cone. By utilizing calculus, specifically integration, we sum up infinitely small volumes to find the total volume. This powerful method allows us to deal with shapes like cones, which have non-uniform distribution of matter along their dimensions. Calculus, through its tools and concepts, helps us model and solve real-world problems efficiently.

Method of Cylindrical Shells

The method of cylindrical shells is an integration technique used for finding volumes. It becomes especially useful in scenarios where the object has rotational symmetry, such as a cone. Instead of breaking down the object into simple geometric shapes (like disks or washers), this method considers it as a series of concentric cylindrical shells.
In the case of a cone, each shell represents a thin, hollow cylinder. By revolving a vertical strip of the cone around the y-axis, we form these shells. Each shell has:

• A radius of rotation, denoted as \(x\).
• A height determined by its vertical position \(y = \frac{h}{r}x\).
• A constant thickness, denoted as \(dx\).

By calculating the volume of each shell and integrating from one end of the cone to the other, we arrive at the cone's total volume.

Integration

Integration is the process of summing up parts to find a whole and is a fundamental concept in calculus. In the context of finding the volume of a cone using cylindrical shells, integration involves summing the volumes of infinitesimally small cylindrical shells.
When calculating the volume, each cylindrical shell’s contribution is expressed as:

• Volume of a single shell: \(dV = \frac{2\pi hx^2}{r} dx\)

This expression allows us to integrate over the entire range of the cone’s base, from \(x = 0\) to \(x = r\). The integration of these incrementally small volumes ultimately gives the full cone volume. This powerful part of calculus lets us calculate complex shapes’ volumes with precision.

Geometric Solids

Geometric solids are fundamental, three-dimensional shapes with defined boundaries like cubes, spheres, and cones. Each solid has characteristic properties, such as surface area and volume. For a cone, specifically a right circular cone, these properties are determined by its height and base radius.
When dealing with a cone, the challenge often lies in deriving and proving properties like volume using mathematical techniques. The volume of a cone is particularly intriguing as it relates directly to the volume of a cylinder. Specifically, a cone’s volume is exactly one third of a cylinder with the same height and base radius, leading to the formula \(V = \frac{1}{3}\pi r^2 h\). Understanding these relationships deepens our comprehension of how geometric solids function and interact with mathematical principles.