Problem 95 - Algebra for College Students

Question

Set up an equation and solve each of the following problems. Suppose that the volume of a sphere is numerically equal to twice the surface area of the sphere. Find the length of a radius of the sphere.

Step-by-Step Solution

Answer

The radius of the sphere is 6 units.

Step 1: Understand the Problem

We need to find the radius of a sphere where its volume equals twice its surface area. The formulas for the volume of a sphere is \( V = \frac{4}{3}\pi r^3 \) and for the surface area is \( A = 4\pi r^2 \). We need to set up an equation with these expressions.

Step 2: Set Up the Equation

According to the problem, \( V = 2A \). Substitute the expressions for volume and surface area: \( \frac{4}{3}\pi r^3 = 2(4\pi r^2) \). This simplifies to \( \frac{4}{3}\pi r^3 = 8\pi r^2 \).

Step 3: Simplify the Equation

Divide both sides by \( \pi \), assuming \( \pi eq 0 \), to get \( \frac{4}{3}r^3 = 8r^2 \). Next, divide both sides by \( r^2 \) to obtain \( \frac{4}{3}r = 8 \).

Step 4: Solve for the Radius

Multiply both sides by \( \frac{3}{4} \) to isolate \( r \): \( r = 8 \times \frac{3}{4} = 6 \).

Step 5: Verify the Solution

To ensure the correctness, substitute \( r = 6 \) back into the expressions for volume and surface area. Compute \( V = \frac{4}{3} \pi (6)^3 \) and \( A = 4 \pi (6)^2 \), then verify that \( V = 2A \). This indeed holds true, so the solution is verified.

Key Concepts

Volume of a Sphere

The volume of a sphere is a measure of the space it occupies, and understanding its computation is crucial in solving problems related to spherical objects. The formula used to calculate the volume is:

• \[ V = \frac{4}{3}\pi r^3 \]

Here, \( V \) represents the volume, \( \pi \) is a constant (approximated to 3.14159), and \( r \) is the radius of the sphere.
This formula illustrates that the volume is proportional to the cube of the radius, implying that even small changes in the radius can lead to significant changes in the volume.
Understanding this principle helps in knowing why larger spheres have exponentially more volume than smaller ones.
To find the volume effectively, ensure to use the correct unit of measurement for the radius, which will also be the unit for the volume (cubed). This concept is essential when comparing sphere sizes or predicting how a change in radius might affect the sphere's capacity.

Surface Area of a Sphere

The surface area of a sphere is a measure of the outer layer of the sphere. It is significant in various applications, like determining the amount of material needed to cover a spherical object. The formula for calculating the surface area is:

• \[ A = 4\pi r^2 \]

In this formula, \( A \) is the surface area, \( \pi \) remains the constant value, and \( r \) is the radius of the sphere.
Just like with volume, the surface area depends heavily on the radius but in a quadratic relation.
This quadratic nature means that doubling the radius will quadruple the surface area.
When solving problems involving surface area, make sure to convert measurements to the same unit to maintain consistency in the units of the area (squared).
Understanding this formula is helpful for practical applications such as painting or coating a spherical object, where knowing the precise surface area determines the amount of material required.

Radius Calculation

Calculating the radius of a sphere is frequently required in geometry when the volume or surface area is provided. To find the radius from the volume or surface area, it involves algebraic manipulation.
In our problem, knowing the relationship \( V = 2A \) gives us the means to set up equations to find the radius.For the given problem:

• Start with the known formulas for volume and surface area: \[ V = \frac{4}{3}\pi r^3 \] \[ A = 4\pi r^2 \]
• Set the problem's condition: \[ \frac{4}{3}\pi r^3 = 8\pi r^2 \]
• Simplify step-by-step to isolate \( r \). First, divide by \( \pi \), and then by \( r^2 \) assuming \( r eq 0 \).This results in: \[ \frac{4}{3}r = 8 \]
• Finally, solve for \( r \): \[ r = 8 \times \frac{3}{4} = 6 \]

This step-by-step algebraic simplification leads us to find the radius.
Having a clear method for calculating the radius is key in solving and understanding these types of geometry problems.

Algebraic Problem Solving

Algebraic manipulation is essential when working on problems where relationships between geometric properties are involved. In this example, it enabled us to solve an equation where a sphere's volume and surface area share a specified relationship.
Here are tips on Algebraic Problem Solving:

• Start by fully understanding the given problem and identifying what's needed to be found.
• Translate the word problem into mathematical equations. Use known formulas and relate them to each other based on the problem's conditions.
• Apply algebraic techniques such as factoring, simplifying, and using inverse operations strategically to isolate the desired variable.
• Check the solution by substituting it back into the original equations to verify accuracy.

Algebra helps in breaking down complex problems into manageable steps, allowing for precise solutions, as seen in the original exercise.
Mastering these skills not only aids in math but enhances problem-solving capabilities broadly across different disciplines.