Problem 49 - 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. \(d=6.4\) meters

Step-by-Step Solution

Answer

Volume is 137.3 cubic meters; surface area is 128.7 square meters.

Step 1: Calculate the Radius

Given the diameter, calculate the radius using the formula for the diameter of a sphere, which is twice the radius. \[ r = \frac{d}{2} \]Substituting the provided diameter of 6.4 meters:\[ r = \frac{6.4}{2} = 3.2 \] meters.

Step 2: Calculate the Volume of the Sphere

Use the formula for the volume of a sphere:\[ V = \frac{4}{3} \pi r^3 \]Substitute the radius we found:\[ V = \frac{4}{3} \pi (3.2)^3 \]Calculate the cube of the radius:\[ 3.2^3 = 32.768 \]So,\[ V = \frac{4}{3} \pi (32.768) \approx 137.258 \text{ cubic meters} \]Approximating to the nearest tenth, the volume is 137.3 cubic meters.

Step 3: Calculate the Surface Area of the Sphere

Use the formula for the surface area of a sphere:\[ A = 4 \pi r^2 \]Substitute the radius:\[ A = 4 \pi (3.2)^2 \]Calculate the square of the radius:\[ 3.2^2 = 10.24 \]So,\[ A = 4 \pi (10.24) \approx 128.678 \text{ square meters} \]Approximating to the nearest tenth, the surface area is 128.7 square meters.

Key Concepts

Understanding Geometry in Terms of Spheres

Geometry is a branch of mathematics that deals with shapes, sizes, and spaces. When it comes to spheres, they are three-dimensional objects defined by points equidistant from a center. Imagine a ball; every point on the surface is the same distance from its middle.

Spheres play a significant role in geometry because they are among the most symmetrical objects studied. Their key features include:

• Center: The midpoint from which every surface point is equidistant.
• Radius (\( r \)): The fixed distance from the center to any point on the sphere.
• Diameter (\( d \)): A straight line passing through the center, connecting two points on the surface. It is always twice the radius (\( d = 2r \)).

Understanding these concepts is crucial for applying formulas to calculate properties like volume and surface area.

Formulas for Volume and Surface Area

To fully understand the spatial properties of spheres, we rely on specific mathematical formulas. These formulas help in determining two primary characteristics: Volume and Surface Area.

For a sphere of radius \(r\):

• Volume (\(V\)): The space inside the sphere is calculated by the formula: \[ V = \frac{4}{3} \pi r^3 \]
• Surface Area (\(A\)): The total area covering the sphere's surface is given by: \[ A = 4 \pi r^2 \]

In these formulas, \(\pi\) is a constant approximately equal to 3.14159.

These formulas are powerful tools in geometry, allowing us to grasp how a sphere's size changes as its radius changes. For example, increasing the radius results in a much larger volume, as shown through the \(r^3\) term in the volume equation.

Mathematical Calculations Through Examples

Calculating a sphere's volume and surface area becomes straightforward when breaking it down step by step. Let’s use an example where the diameter of a sphere is 6.4 meters.

First, we determine the radius by dividing the diameter by 2: \[ r = \frac{6.4}{2} = 3.2 \] meters.

Next, to find the volume, apply the volume formula:\[ V = \frac{4}{3} \pi (3.2)^3 \] Calculate \(3.2\) to the power of 3 to get 32.768. Then the volume is approximately:\[ 137.3 \text{ cubic meters} \] when approximated to the nearest tenth.

For the surface area, use:\[ A = 4 \pi (3.2)^2 \] Calculate \(3.2\) squared to get 10.24. So, the surface area is:\[ 128.7 \text{ square meters} \], also rounded to the nearest tenth.

Practicing these calculations reinforces how mathematical principles apply to real-world objects, enhancing your understanding of geometry and improving problem-solving skills.