Problem 31 - Algebra 2

Question

Geometry The formula for the volume of a sphere is \(V=\frac{4}{3} \pi r^{3} .\) Find the radius to the nearest hundredth of a sphere with each volume. $$ 0.45 \mathrm{cm}^{3} $$

Step-by-Step Solution

Answer

The radius \( r \) of the sphere is approximately 0.48 cm when rounded off to the nearest hundredth.

Step 1: Understand and Rearrange the Volume Formula

The formula for the volume of a sphere is given by \( V= \frac{4}{3} \pi r^{3} \). We are supposed to find the radius \( r \), meaning we need to rearrange the formula to make \( r \) the subject. That formula is \( r = \left( \frac{3V}{4\pi} \right) ^{\frac{1}{3}} \)

Step 2: Substitute the Given Volume

The given volume of the sphere \( V \) is 0.45 cubic cm. Substituting this value into the rearranged formula we get: \( r = \left( \frac{3*0.45}{4\pi} \right) ^{\frac{1}{3}} \)

Step 3: Solve for the Radius

Now we just have to do the calculations on the right-hand side of the equation which should give us the radius \( r \). Remember to round off your answer to the nearest hundredth as prompted by the question

Key Concepts

Calculating Volume

Calculating the volume of a sphere involves applying a specific formula. This measurement is the amount of space the sphere occupies in three-dimensional space.
The formula used for this is:

• \[ V = \frac{4}{3} \pi r^3 \]

Here, **\(V\)** represents the volume, **\(\pi\)** is approximately 3.14159, and **\(r\)** is the radius of the sphere. This formula tells us that the volume is derived by raising the radius to the third power and then using the constant factors \(\frac{4}{3}\) and \(\pi\). The sphere's radius is a critical component in determining how much space it fills.
When given the volume, our task is sometimes to manipulate this formula to find the unknown radius.

Geometry Formulas

Geometry provides us with various formulas to understand and solve for properties of different shapes and figures. For spheres, the volume and surface area are key characteristics often calculated. In our case, the formula for volume is central.
Why does this formula work? It stems from integral calculus, capturing the idea of building the volume from many infinitesimally small discs stacked within the sphere. These formulas support solving real-world problems involving capacity and storage.
Knowing these calculus-based origins isn't just academic; it helps students appreciate the logical foundation underlying geometry definitions and calculations.
Understanding and being able to apply different geometry formulas is essential, simplifying complex relationships into comprehensible steps.

Solving for a Variable

To solve for a variable like the radius in our volume formula, we need to rearrange the equation so that the variable is isolated.
Initially, our sphere's volume formula is:

• \[ V = \frac{4}{3} \pi r^3 \]

If the goal is to find **\(r\)**, rearrange this equation as follows:

• Multiply both sides by \(\frac{3}{4\pi}\), which gives \( r^3 = \frac{3V}{4\pi} \)
• Take the cube root of both sides, resulting in \( r = \left( \frac{3V}{4\pi} \right) ^{\frac{1}{3}} \)

This process involves algebraic manipulation, ensuring that all operations on one side of the equation are balanced by the other side.
Successfully solving for a variable strengthens your problem-solving skills, helping you tackle broader mathematical challenges.