Problem 65 - Calculus with Concepts in Calculus

Question

A sphere with surface area \(s\) has a radius \(r(s)\) given by $$ r(s)=\frac{1}{2} \sqrt{\frac{s}{\pi}} $$ a. Using the formula in Exercise 64 , find a formula for the volume of a sphere in terms of its surface area. b. Determine the volume corresponding to a surface area of 6 .

Step-by-Step Solution

Answer

The volume formula in terms of surface area is \( V = \frac{1}{6} \sqrt{\frac{s^3}{\pi}} \). For \( s = 6 \), the volume is approximately 2.194.

Step 1: Recall the formula for the volume of a sphere

The volume \( V \) of a sphere with radius \( r \) is given by the formula: \[ V = \frac{4}{3} \pi r^3 \]

Step 2: Substitute the expression for radius

Using the given expression for the radius \( r(s) = \frac{1}{2} \sqrt{\frac{s}{\pi}} \), substitute \( r(s) \) into the volume formula:\[ V = \frac{4}{3} \pi \left( \frac{1}{2} \sqrt{\frac{s}{\pi}} \right)^3 \]

Step 3: Simplify the expression

Simplify the expression inside the volume formula:- Cube the expression: \[ \left( \frac{1}{2} \sqrt{\frac{s}{\pi}} \right)^3 = \frac{1}{8} \left( \frac{s}{\pi} \right)^{3/2} \] - Substitute back into the volume formula: \[ V = \frac{4}{3} \pi \times \frac{1}{8} \left( \frac{s}{\pi} \right)^{3/2} \]- Simplify further: \[ V = \frac{1}{6} \sqrt{\frac{s^3}{\pi}}\]

Step 4: Compute the volume for a given surface area

Using the derived volume formula \( V = \frac{1}{6} \sqrt{\frac{s^3}{\pi}}\), plug in \( s = 6 \):\[ V = \frac{1}{6} \sqrt{\frac{6^3}{\pi}} \]

Step 5: Calculate the numeric result

Perform the arithmetic:- Calculate \( 6^3 = 216 \).- Substitute into the expression: \[ V = \frac{1}{6} \sqrt{\frac{216}{\pi}} \]- Compute it to find: \[ V \approx 2.194 \text{ (rounded to three decimal places)} \]

Key Concepts

Sphere Volume Formula

The sphere volume formula is a fundamental concept in geometry. It helps us determine how much space is inside a sphere. The traditional formula to calculate the volume, \( V \), of a sphere with a radius \( r \) is:

• \( V = \frac{4}{3} \pi r^3 \)

This formula uses the constant \( \pi \), approximately 3.1416, which is crucial since a sphere is a three-dimensional shape.
Understanding the sphere volume formula is key to solving many problems related to spheres in calculus and geometry. It provides insights into how varying the radius affects the volume of the sphere.
When the radius increases, the volume does, too, and to a significant degree because of the cube exponent.

Surface Area to Volume Conversion

Converting a sphere's surface area to its volume can be an intriguing mathematical task. To perform this conversion, you need to understand the relationship between a sphere's surface area and its volume.
The surface area of a sphere is calculated with the formula:

• \( A = 4 \pi r^2 \)

In our problem, we used a variation by expressing the radius \( r \) in terms of the surface area \( s \):

• \( r = \frac{1}{2} \sqrt{\frac{s}{\pi}} \)

Then, this expression is replaced in the volume formula. This substitution allows us to express volume in terms of surface area, yielding an alternate formula for volume:

• \( V = \frac{1}{6} \sqrt{\frac{s^3}{\pi}} \)

By converting from surface area to volume, you engage a geometric concept that reveals the spatial relationship of a sphere.

Radius of Sphere Formula

Finding the radius of a sphere from its surface area involves some mathematical manipulations. The radius formula found in this exercise is:

• \( r(s) = \frac{1}{2} \sqrt{\frac{s}{\pi}} \)

This equation shows how you can derive the radius \( r \) using surface area \( s \). Understanding this radius formula is particularly useful when you encounter problems where surface area is known but radius is not.
It simplifies the process because you don't have to directly measure the radius, which might be impractical in real-world scenarios. This formula directly supports the connection between different measurements of spherical objects.
It's also a great example of using algebraic methods to transform and manipulate equations for use in solving more complex problems.

Mathematical Simplification Steps

Simplifying mathematical expressions is a valuable skill in calculus and helps to transform complex equations into more manageable forms. In our exercise, the simplification process involves multiple steps:

• Substitute the radius formula into the volume equation.
• Simplify the terms inside the parentheses before cubing them.
• Cube the expression for radius, leading to: \( \left( \frac{1}{2} \sqrt{\frac{s}{\pi}} \right)^3 = \frac{1}{8} \left( \frac{s}{\pi} \right)^{3/2} \)
• Finally, substitute back into the volume equation and simplify further.
• This results in \( V = \frac{1}{6} \sqrt{\frac{s^3}{\pi}} \), a single expression for volume based on surface area.

Each of these steps is methodical, illustrating algebraic manipulation principles essential in calculus. The key is to break down complex tasks into individual, simpler steps to make the problem-solving process more achievable.