Problem 50 - College Algebra with Modeling and Visualization

Question

Find the volume and surface area of the sphere satisfying the given condition, where \(r\) is the radius and dis the diameter. Approximate values to the nearest tenth. \(d=16\) feet

Step-by-Step Solution

Answer

Volume: 682.7 cubic feet; Surface Area: 804.2 square feet.

Step 1: Establish Relationship Between Diameter and Radius

The radius of a sphere, denoted by \( r \), is half of its diameter. Given that the diameter \( d = 16 \) feet, the radius is calculated as \( r = \frac{d}{2} = \frac{16}{2} = 8 \) feet.

Step 2: Calculate the Volume of the Sphere

The formula for the volume \( V \) of a sphere is \( V = \frac{4}{3} \pi r^3 \). Substituting \( r = 8 \) feet into the formula, we have:\[V = \frac{4}{3} \pi (8)^3 = \frac{4}{3} \pi \times 512 \approx \frac{2048}{3} \pi \approx 682.7\ \text{cubic feet}.\]

Step 3: Calculate the Surface Area of the Sphere

The surface area \( A \) of a sphere is given by the formula \( A = 4 \pi r^2 \). Substituting \( r = 8 \) feet, we calculate:\[A = 4 \pi (8)^2 = 4 \pi \times 64 = 256 \pi \approx 804.2\ \text{square feet}.\]

Step 4: Finalize the Approximate Values

The calculated volume of the sphere is approximately 682.7 cubic feet, and the surface area is approximately 804.2 square feet.

Key Concepts

Volume of a Sphere

To find the volume of a sphere, which is the space it occupies, we use a specific mathematical formula. The volume formula is expressed as \[V = \frac{4}{3} \pi r^3\] where:

• \( V \) is the volume,
• \( \pi \) is a mathematical constant approximately equal to 3.14159,
• \( r \) is the radius of the sphere.

In our exercise, we need the radius, which we get by halving the diameter. Given that the diameter is 16 feet, the radius \( r \) is 8 feet. Plugging this radius into the formula gives us the volume. This approach helps us to understand how alterations in the radius can affect the volume of a sphere significantly.

It's important to note that the volume increases with the cube of the radius, meaning that a small change in radius leads to a large change in volume. This is why spheres play a crucial role in various fields, from physics to design, where maximizing or minimizing space is important.

Surface Area of a Sphere

The surface area of a sphere is the total area that covers its outer layer. Calculating this can sometimes seem daunting, but with the formula, it's straightforward:\[A = 4 \pi r^2\] where:

• \( A \) is the surface area,
• \( \pi \) is again our constant, approximately 3.14159,
• \( r \) is the radius of the sphere.

By knowing the radius is 8 feet, we substitute this value into the formula to calculate the surface area. This gives us the overall coverage of the sphere, which is essential for many practical applications, like determining the material needed to cover a spherical object.

The surface area relates to the square of the radius, indicating that as the radius grows, the surface area increases dramatically. This concept is useful in fields that work with surface phenomena, including meteorology and materials science.

Mathematical Formulas

Mathematical formulas are key tools in solving problems related to spheres. They transform complex real-world scenarios into solvable mathematical equations. Understanding how to apply these formulas effectively is essential.
By breaking down real-life objects into mathematical components, such as using a sphere as a model, we employ formulas to calculate characteristics like volume and surface area. Each variable and constant in these formulas serves a purpose and understanding them ensures accurate computation.
Linear relationships, like the one between the diameter and radius where \( r = \frac{d}{2} \), are also crucial. This conversion is often the first step in applying more complex formulas like those for volume and surface area. It underscores the importance of mastering basic algebraic manipulations when working with geometric shapes.
Overall, formulas provide a framework, allowing us to delve into deeper mathematical analysis and harness these insights in practical and theoretical applications.