Problem 48
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. \(r=4.1\) inches
Step-by-Step Solution
Verified Answer
The volume is 288.7 cubic inches, and the surface area is 211.2 square inches.
1Step 1: Understand the Formulas
The volume of a sphere is calculated with the formula \( V = \frac{4}{3} \pi r^3 \). The surface area of a sphere is given by \( A = 4 \pi r^2 \). We'll use these formulas to find the volume and surface area, given \( r = 4.1 \) inches.
2Step 2: Calculate the Volume
Substitute \( r = 4.1 \) into the volume formula: \[ V = \frac{4}{3} \pi (4.1)^3 \]. First, compute \( (4.1)^3 = 68.921 \). Then, \[ V = \frac{4}{3} \pi \times 68.921 \approx 288.707 \text{ cubic inches} \]. Rounding to the nearest tenth gives a volume of 288.7 cubic inches.
3Step 3: Calculate the Surface Area
Substitute \( r = 4.1 \) into the surface area formula: \[ A = 4 \pi (4.1)^2 \]. Compute \( (4.1)^2 = 16.81 \). Then, \[ A = 4 \pi \times 16.81 \approx 211.221 \text{ square inches} \]. Rounding to the nearest tenth, the surface area is 211.2 square inches.
Key Concepts
SphereRadius and DiameterMathematical Formulas
Sphere
A sphere is a perfectly round three-dimensional shape, much like a ball. It is unique in geometry because every point on its surface is the same distance from its center. This distance is what we call the radius of the sphere.
Spheres are fascinating because of their symmetry and uniformity. There's no flat surface on a sphere, and every line through the center bisects it, creating symmetrical halves. Understanding this fundamental shape is essential for tasks involving volume and surface area for objects like planets, balls, and bubbles. When working with spheres, remember that all dimensions radiate out from the center, creating a harmonious form.
Spheres are fascinating because of their symmetry and uniformity. There's no flat surface on a sphere, and every line through the center bisects it, creating symmetrical halves. Understanding this fundamental shape is essential for tasks involving volume and surface area for objects like planets, balls, and bubbles. When working with spheres, remember that all dimensions radiate out from the center, creating a harmonious form.
Radius and Diameter
In a sphere, the radius is a crucial measurement. The radius ( \( r \)) is the distance from the center of the sphere to any point on its surface. Knowing this allows us to calculate many important properties of the sphere.
The diameter ( \( d \)) is another vital term. It represents the longest distance across the sphere, passing through the center. Formula-wise, the diameter is simply twice the radius: \( d = 2r \). A clear understanding of these terms is helpful because many problems and formulas will refer to them.
Always ensure to identify if you're working with the radius or diameter before using any mathematical formulas to avoid mistakes.
The diameter ( \( d \)) is another vital term. It represents the longest distance across the sphere, passing through the center. Formula-wise, the diameter is simply twice the radius: \( d = 2r \). A clear understanding of these terms is helpful because many problems and formulas will refer to them.
Always ensure to identify if you're working with the radius or diameter before using any mathematical formulas to avoid mistakes.
Mathematical Formulas
Mathematical formulas related to spheres help us calculate quantities like volume and surface area. Knowing these formulas enables you to solve problems efficiently.
The Volume \( V \) of a sphere is found using: \( V = \frac{4}{3} \pi r^3 \). This formula calculates how much space the sphere occupies. For example, with \( r = 4.1 \) inches, substituting gives the volume, which when calculated and approximated, results in \( 288.7 \) cubic inches.
The Surface Area \( A \) is derived using: \( A = 4 \pi r^2 \). It represents the total area that covers the Sphere's surface. With the same radius of \( 4.1 \) inches, the surface area calculates and rounds to about \( 211.2 \) square inches.
These formulas are essential tools for various applications, from engineering to everyday problem-solving, where spherical shapes are involved. Remember these key formulas for effective calculations!
The Volume \( V \) of a sphere is found using: \( V = \frac{4}{3} \pi r^3 \). This formula calculates how much space the sphere occupies. For example, with \( r = 4.1 \) inches, substituting gives the volume, which when calculated and approximated, results in \( 288.7 \) cubic inches.
The Surface Area \( A \) is derived using: \( A = 4 \pi r^2 \). It represents the total area that covers the Sphere's surface. With the same radius of \( 4.1 \) inches, the surface area calculates and rounds to about \( 211.2 \) square inches.
These formulas are essential tools for various applications, from engineering to everyday problem-solving, where spherical shapes are involved. Remember these key formulas for effective calculations!
Other exercises in this chapter
Problem 48
Write the expression in radical notation. Then evaluate the expression when the result is an integer. $$ 8^{4 / 3} $$
View solution Problem 48
Simplify the expression. $$ \frac{x^{2}+x-12}{2 x^{2}-9 x-5} \div \frac{x^{2}+7 x+12}{2 x^{2}-7 x-4} $$
View solution Problem 49
Apply the distributive property. $$-2(5 x+9)$$
View solution Problem 49
Exercises \(45-54:\) Use the quotient rule to simplify the expression. Use positive exponents to write your answer. $$ \frac{24 x^{3}}{6 x} $$
View solution