Problem 35 - Algebra and Trigonometry

Question

Write an algebraic formula for the given quantity. You may need to consult the formulas for area and volume listed on the inside front cover of this book. The volume \(V\) of what remains of a solid ball of radius \(R\) after a smaller solid ball of radius \(r\) is removed from inside the larger ball, as shown in the figure

Step-by-Step Solution

Answer

The volume of the remaining part is \( V = \frac{4}{3} \pi (R^3 - r^3) \).

Step 1: Understanding the Problem

We are given a larger solid ball with a smaller ball removed from its interior, and we need to find the formula for the volume of the remaining part. The larger ball has radius \(R\) and the smaller ball has radius \(r\).

Step 2: Formula for Volume of a Sphere

The formula for the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \). We will use this formula to find the volumes of both the larger and smaller spheres.

Step 3: Volume of the Larger Sphere

Apply the formula for the volume of a sphere to the larger sphere. The volume \( V_{\text{large}} \) is \( V_{\text{large}} = \frac{4}{3} \pi R^3 \).

Step 4: Volume of the Smaller Sphere

Apply the same formula to find the volume of the smaller sphere. The volume \( V_{\text{small}} \) is \( V_{\text{small}} = \frac{4}{3} \pi r^3 \).

Step 5: Calculating the Remaining Volume

To find the volume of the remaining part of the solid ball, subtract the volume of the smaller sphere from the volume of the larger sphere: \( V = V_{\text{large}} - V_{\text{small}} = \frac{4}{3} \pi R^3 - \frac{4}{3} \pi r^3 \).

Step 6: Factor Out the Common Term

Factor out the common term \( \frac{4}{3} \pi \) from the expression: \( V = \frac{4}{3} \pi (R^3 - r^3) \). This expression gives the volume of the remaining part of the solid ball.

Key Concepts

Algebraic Formula

The core of solving volume problems in solid geometry often starts with understanding algebraic formulas. An algebraic formula in this context is an equation that expresses a physical quantity, like volume, in terms of variables.
In the case of a sphere, the volume is represented by the formula \( V = \frac{4}{3} \pi r^3 \). Here, \( V \) is the volume, \( \pi \) is a constant (about 3.14159), and \( r \) is the radius of the sphere.
This formula is derived using integral calculus to account for the infinite number of infinitesimally thin circular disks that compose the sphere.Algebraic formulas help simplify complex geometrical relationships by providing a direct calculation method using one or more variables. They turn a visual 3D problem into a manageable calculation process.

Volume Calculation

Calculating the volume of solid objects is a fundamental aspect of geometry. To find the volume of complex shapes, we often break them down into simpler components with known formulas.

For spheres, as mentioned, we use the formula \( V = \frac{4}{3} \pi r^3 \). When dealing with two spheres, as in our exercise, calculate their respective volumes separately:

• The larger sphere with radius \( R \) has a volume of \( V_{\text{large}} = \frac{4}{3} \pi R^3 \).
• The smaller sphere with radius \( r \) has a volume of \( V_{\text{small}} = \frac{4}{3} \pi r^3 \).

The problem requires the volume of a composite shape (the larger sphere minus the smaller one). This is accomplished by subtracting the volume of the smaller sphere from the larger one:\[ V = \frac{4}{3} \pi (R^3 - r^3) \] This subtraction represents the remaining volume after removing the smaller sphere from the larger sphere.

Solid Geometry

Solid geometry is the study of three-dimensional figures like spheres, cubes, and pyramids. It often involves determining volumes, surface areas, and other properties specific to three-dimensional objects.
Spheres, for instance, are a key focus due to their natural occurrence and simple symmetrical properties. In our problem, we explore spheres by examining the concept of one sphere within another. Understanding solid geometry requires familiarity with the properties of basic shapes. For spheres, this includes knowing how to use the radius to compute both surface area and volume.
Additionally, solid geometry helps solve real-world problems, such as calculating the amount of material needed to produce hollow objects or finding leftover spaces when one object is removed from another. Utilizing geometrical formulas, like those for spheres, enables practical applications and a deeper understanding of spatial relationships.