Chapter 1: Problem 4
the volume \(V\) of a sphere as a function of the sphere's radius \(r ;\) the volume of a sphere of radius 3 cm.
Short Answer
Expert verified
The volume of the sphere is approximately 113.097 cubic cm.
Step by step solution
01
Understand the formula
The formula for the volume of a sphere is given by \(V = \frac{4}{3} \pi r^3\), where \(V\) is the volume and \(r\) is the radius of the sphere. In this exercise, we are asked to find the volume of a sphere with radius 3 cm.
02
Insert radius into the formula
Now that we know the radius of the sphere, we can replace \(r\) in the formula with 3. This gives us \(V = \frac{4}{3} \pi (3)^3\).
03
Perform calculations
Next, perform the multiplication and calculate the volume. \(V=\frac{4}{3} \pi (27)\), which simplifies to \(V = 36\pi\) cubic cm.
04
Simplify the answer to decimal form
As a last step, to obtain a numerical result, let's approximate \(\pi\) by 3.14 and multiply it with 36. This gives a volume of approximately 113.097 cubic cm.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Sphere Volume Formula
Grasping the concept of the sphere volume formula is essential in geometry and various practical applications. The volume (\(V\)) of a sphere with radius (\(r\)) is given by the formula \(V = \frac{4}{3} \pi r^3\). This formula comes from the mathematical constant \(\pi\), which is roughly 3.14159 and represents the ratio of the circumference of a circle to its diameter.
When solving problems involving the volume of a sphere, it is important to remember that the radius is the distance from the center of the sphere to any point on its surface. The exponent 3 in the formula indicates that the radius is cubed, thus considering all three dimensions that contribute to the sphere's volume. For example, for a sphere with a radius of 3 cm, the formula becomes \(V = \frac{4}{3} \pi (3)^3\), leading to \(V = 36\pi\) cubic cm when you perform the algebraic manipulation of raising 3 to the third power and multiplying by \(\pi\) and \(4/3\).
To compute the exact volume, the last step involves converting \(\pi\) to its decimal approximation (3.14) and multiplying it with the rest of the numbers to arrive at the solution.
When solving problems involving the volume of a sphere, it is important to remember that the radius is the distance from the center of the sphere to any point on its surface. The exponent 3 in the formula indicates that the radius is cubed, thus considering all three dimensions that contribute to the sphere's volume. For example, for a sphere with a radius of 3 cm, the formula becomes \(V = \frac{4}{3} \pi (3)^3\), leading to \(V = 36\pi\) cubic cm when you perform the algebraic manipulation of raising 3 to the third power and multiplying by \(\pi\) and \(4/3\).
To compute the exact volume, the last step involves converting \(\pi\) to its decimal approximation (3.14) and multiplying it with the rest of the numbers to arrive at the solution.
Geometric Applications of Integration
Integration is a fundamental tool in calculus, and it has a special place when it comes to finding volumes of solids with curved surfaces, like spheres. In geometric applications, integration is used to sum up an infinite number of infinitesimally thin slices to calculate the total volume of a three-dimensional object.
This technique is particularly helpful for objects whose cross-sectional area changes at different points along a certain dimension. For a sphere, integration can be visualized by imagining the sphere sliced into numerous thin layers. Each layer is a disk with an area that can be expressed as a function of its radius. By integrating this disk area function across the sphere’s diameter, the sphere’s volume can be computed – which, as seen in exercises, results in the sphere volume formula \(V = \frac{4}{3} \pi r^3\).
Understanding the relationship between integration and volume can deepen students' comprehension of how calculus is used to solve real-world problems that involve complex shapes. It empowers them to tackle a wide array of problems beyond those with straightforward geometric figures.
This technique is particularly helpful for objects whose cross-sectional area changes at different points along a certain dimension. For a sphere, integration can be visualized by imagining the sphere sliced into numerous thin layers. Each layer is a disk with an area that can be expressed as a function of its radius. By integrating this disk area function across the sphere’s diameter, the sphere’s volume can be computed – which, as seen in exercises, results in the sphere volume formula \(V = \frac{4}{3} \pi r^3\).
Understanding the relationship between integration and volume can deepen students' comprehension of how calculus is used to solve real-world problems that involve complex shapes. It empowers them to tackle a wide array of problems beyond those with straightforward geometric figures.
Algebraic Manipulation
Algebraic manipulation is the process of rearranging and simplifying mathematical expressions and equations. It is a skill that is foundational in solving geometry problems, like finding the volume of a sphere. Properly applying algebraic techniques enables one to evaluate expressions and formulas accurately.
When working with the sphere volume formula, algebraic manipulation includes operations such as cubing the radius, multiplying by \(\pi\), and multiplying by \(4/3\). Each step systematically reduces the complexity of the original expression, eventually translating it into an easily understandable numerical form. For instance, in our volume calculation for a sphere with a 3 cm radius, we first cube 3 to get 27, multiply that by \(\pi\), and then multiply by \(4/3\), which we can further simplify into \(36\pi\) cm^3.
Moreover, understanding algebraic manipulation can also help students with the reverse process: given the volume, working back to find other sphere characteristics, such as radius or diameter. Therefore, a solid grasp of algebraic manipulation is crucial for a comprehensive understanding of geometric formulas and their applications.
When working with the sphere volume formula, algebraic manipulation includes operations such as cubing the radius, multiplying by \(\pi\), and multiplying by \(4/3\). Each step systematically reduces the complexity of the original expression, eventually translating it into an easily understandable numerical form. For instance, in our volume calculation for a sphere with a 3 cm radius, we first cube 3 to get 27, multiply that by \(\pi\), and then multiply by \(4/3\), which we can further simplify into \(36\pi\) cm^3.
Moreover, understanding algebraic manipulation can also help students with the reverse process: given the volume, working back to find other sphere characteristics, such as radius or diameter. Therefore, a solid grasp of algebraic manipulation is crucial for a comprehensive understanding of geometric formulas and their applications.
