Chapter 12: Problem 14
Find the volume of a sphere with a radius of 10 inches. Round to the nearest hundredth.
Short Answer
Expert verified
The volume of the sphere is approximately 4188.79 cubic inches.
Step by step solution
01
Understanding the Formula
The formula to find the volume of a sphere is given by \( V = \frac{4}{3} \pi r^3 \). Here, \( V \) represents the volume, \( r \) is the radius, and \( \pi \) is approximately 3.14159.
02
Substitute the Radius
In this exercise, the sphere's radius \( r \) is given as 10 inches. Substitute \( r = 10 \) into the volume formula: \( V = \frac{4}{3} \pi (10)^3 \).
03
Calculate the Cube of the Radius
Calculate \( 10^3 \), which is equal to 1000. So, the equation now is \( V = \frac{4}{3} \pi \times 1000 \).
04
Multiply by Pi
Multiply \( 1000 \times \pi \) which is approximately \( 1000 \times 3.14159 = 3141.59 \).
05
Apply the Formula Division
Now multiply \( \frac{4}{3} \times 3141.59 \), which is approximately \( 4188.79 \).
06
Round to Nearest Hundredth
The calculated volume of the sphere is approximately 4188.79. Since it is already to the nearest hundredth, no further rounding is necessary.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Geometry Formula
When calculating the volume of a sphere, one of the fundamental tools you'll use is the geometry formula for sphere volume. The formula is expressed as \( V = \frac{4}{3} \pi r^3 \). In this equation, \( V \) stands for volume, \( \pi \) is a mathematical constant approximately equal to 3.14159, and \( r \) symbolizes the radius of the sphere.
This particular formula is derived from using integral calculus and is based on the symmetrical nature of a sphere. Understanding each component of the formula is crucial. The \( \frac{4}{3} \) comes from integrating the surface area of a sphere over its radius.
Knowing this formula allows you to calculate how much space a sphere occupies, which is extremely valuable in various scientific fields, including physics and engineering. It's essential to memorize and understand it, as it's the first step towards mastering three-dimensional geometry.
For better retention, remember that this formula requires you to cube the radius (\( r^3 \)), emphasizing the impact of the radius on the volume of the sphere. As the radius increases, so does the volume, to a significant extent.
This particular formula is derived from using integral calculus and is based on the symmetrical nature of a sphere. Understanding each component of the formula is crucial. The \( \frac{4}{3} \) comes from integrating the surface area of a sphere over its radius.
Knowing this formula allows you to calculate how much space a sphere occupies, which is extremely valuable in various scientific fields, including physics and engineering. It's essential to memorize and understand it, as it's the first step towards mastering three-dimensional geometry.
For better retention, remember that this formula requires you to cube the radius (\( r^3 \)), emphasizing the impact of the radius on the volume of the sphere. As the radius increases, so does the volume, to a significant extent.
Mathematical Calculation
Performing a mathematical calculation using the volume formula involves several distinct steps that need to be carried out with care. Let's break this down using our exercise as an example.
Firstly, you start by substituting the value of the sphere's radius into the formula. For a radius of 10 inches, your equation becomes \( V = \frac{4}{3} \pi (10)^3 \).
The next step is to calculate the cube of the radius. Cubing the radius, we compute \( 10^3 = 1000 \). Insert this value back into the formula resulting in \( V = \frac{4}{3} \pi \times 1000 \).
Calculations become quite easy once you are familiar with breaking down each step methodically. Now, your task is to multiply \( 1000 \) by \( \pi \), resulting in approximately \( 3141.59 \) (using \( \pi \approx 3.14159 \)).
Finally, this value is multiplied by \( \frac{4}{3} \), leading to a calculated volume of approximately \( 4188.79 \). By following these steps clearly and consistently, you ensure accuracy in your mathematical calculations.
Firstly, you start by substituting the value of the sphere's radius into the formula. For a radius of 10 inches, your equation becomes \( V = \frac{4}{3} \pi (10)^3 \).
The next step is to calculate the cube of the radius. Cubing the radius, we compute \( 10^3 = 1000 \). Insert this value back into the formula resulting in \( V = \frac{4}{3} \pi \times 1000 \).
Calculations become quite easy once you are familiar with breaking down each step methodically. Now, your task is to multiply \( 1000 \) by \( \pi \), resulting in approximately \( 3141.59 \) (using \( \pi \approx 3.14159 \)).
Finally, this value is multiplied by \( \frac{4}{3} \), leading to a calculated volume of approximately \( 4188.79 \). By following these steps clearly and consistently, you ensure accuracy in your mathematical calculations.
Volume Measurement
Volume measurement is essential in numerous real-world applications, including engineering and architecture, where understanding the space occupied by an object is critical. When we talk about the volume of a sphere, we are essentially discussing the amount of space contained within its three-dimensional boundary.
This measurement is different from area, which only looks at two-dimensional space. Volume, being a three-dimensional concept, involves units of cubic measurements. For example, when measuring with a radius in inches, the resulting volume is given in cubic inches.
Volume measurement in spheres, like in our exercise, uses precise mathematical principles. These principles allow us to predict how much substance it can hold or how much space it might displace.
Acquiring a good grasp of measuring volume helps in fields such as fluid dynamics, where scientists calculate how much liquid can be stored within a spherical tank, or in astronomy, where the size of planets is determined. Whether for practical applications or pure curiosity, having an understanding of volume measurement is invaluable in comprehending the physical world.
This measurement is different from area, which only looks at two-dimensional space. Volume, being a three-dimensional concept, involves units of cubic measurements. For example, when measuring with a radius in inches, the resulting volume is given in cubic inches.
Volume measurement in spheres, like in our exercise, uses precise mathematical principles. These principles allow us to predict how much substance it can hold or how much space it might displace.
Acquiring a good grasp of measuring volume helps in fields such as fluid dynamics, where scientists calculate how much liquid can be stored within a spherical tank, or in astronomy, where the size of planets is determined. Whether for practical applications or pure curiosity, having an understanding of volume measurement is invaluable in comprehending the physical world.
