Problem 24 - The Calculus with Analytic Geometry

Question

Use the theorem of Pappus to find the volume of a sphere with a radius of \(r\) units.

Step-by-Step Solution

Answer

The volume of the sphere is \frac{4\pi r^3}{3}.

Step 1: Understand the Theorem of Pappus

Theorem of Pappus states that the volume of a solid of revolution generated by rotating a plane curve about an external axis (that is, an axis not intersecting the curve) is equal to the product of the area of the curve and the distance traveled by the centroid of the curve.

Step 2: Identify the Plane Curve and Axis of Rotation

For a sphere, consider rotating a semicircle of radius \(r\) about its diameter. The semicircle will generate a full sphere when rotated around the diameter.

Step 3: Calculate the Area of the Semicircle

The area of a semicircle is half the area of a full circle. The formula for the area of a full circle is \(\text{A} = \pi r^2\), so the area of the semicircle is \(\text{A}_{\text{semicircle}} = \frac{1}{2} \pi r^2\).

Step 4: Determine the Centroid of the Semicircle

The centroid of a semicircle is located at a distance of \(\frac{4r}{3\pi}\) from the base along the axis of symmetry of the semicircle.

Step 5: Calculate the Distance Traveled by the Centroid

When the semicircle is rotated about its diameter, the centroid travels a circular path with a radius equal to the distance from the centroid to the axis. This distance is \(\frac{4r}{3\pi}\). Therefore, the distance traveled by the centroid is the circumference of this path, which is \(2\pi \left(\frac{4r}{3\pi}\right) = \frac{8r}{3}\).

Step 6: Compute the Volume Using the Theorem of Pappus

According to the theorem of Pappus, the volume \(V\) is the product of the area of the semicircle and the distance traveled by the centroid: \[ V = \left(\frac{1}{2} \pi r^2\right) \left(\frac{8r}{3}\right) = \frac{4\pi r^3}{3} \]. Therefore, the volume of the sphere is \frac{4\pi r^3}{3}.

Key Concepts

Theorem of Pappus

The Theorem of Pappus is a cornerstone in geometric studies. It is essential for understanding volumes of solids created by rotation. The theorem states that the volume of a solid of revolution is equal to the product of the area of the rotating plane curve and the distance traveled by its centroid.
For example, consider a semicircle of radius \(r\) that rotates around its diameter to form a sphere. The area of the semicircle is \(\frac{1}{2} \pi r^2\). The centroid of this semicircle moves in a circular path with a radius \(\frac{4r}{3\pi}\). This knowledge leads us to calculate the distance and volume in further steps. These calculations confirm the Theorem of Pappus and show its practical applications in determining volumes of different shapes.

Solid of Revolution

A solid of revolution is formed by rotating a two-dimensional shape around an axis. This concept helps in visualizing and calculating volumes. To form a sphere from a semicircle, you rotate the semicircle around its diameter. Here’s how:

• A semicircle's base lies on the x-axis.
• When you rotate the semicircle around its diameter (y-axis), it forms a three-dimensional sphere.

The generated volume depends on the initial area and the path traveled by the centroid. Understanding solids of revolution simplifies complex volume calculations by breaking them down into manageable steps. This concept is widely used in engineering and physics to describe rotational bodies.

Centroid

The centroid of a shape is like its center of mass, and it plays a vital role in applying the Theorem of Pappus. For a semicircle, the centroid lies \(\frac{4r}{3\pi}\) units from the flat base. This distance affects the volume calculations significantly.
To find the distance the centroid travels during rotation:

• First, locate the centroid.
• Next, determine the circular path radius which is \(\frac{4r}{3\pi}\).

The total path traveled by this centroid is the circumference: \(2\pi \left(\frac{4r}{3\pi}\right) = \frac{8r}{3}\). By multiplying this distance with the area, you utilize Pappus' theorem to find the volume effectively. Mastering the concept of a centroid simplifies calculations and enhances understanding of geometric properties.