Volume of Sphere

8 Key excerpts on "Volume of Sphere"

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  Elementary Geometry for College Students

  • Daniel C. Alexander, Geralyn M. Koeberlein, , , Daniel C. Alexander, Geralyn M. Koeberlein(Authors)
  • 2014(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  EXS. 8–10 Figure 9.43 THEOREM 9.4.3 The volume V of a sphere with a radius of length r is given by . V 4 3 p r 3 EXAMPLE 4 Find the exact volume of a sphere whose length of radius is 1.5 in. SOLUTION This calculation can be done more easily if we replace 1.5 by . 9 p 2 in 3 4 3 p 3 2 3 2 3 2 V 4 3 p r 3 3 2 D iscover A farmer’s silo is a composite shape. That is, it is actually composed of two solids. What are they? ANSWER Cylinder and hemisphere © James Horning/Shutterstock.com Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. EXAMPLE 5 A spherical propane gas storage tank has a volume of . Using , find the radius of the sphere. SOLUTION becomes . Then . In turn, The radius of the tank is 3 ft. Just as two concentric circles have the same center but different lengths of radii, two spheres can also be concentric. This fact is the basis for the solution of the problem in the following example. EXAMPLE 6 A child’s hollow plastic ball has an inside diameter length of 10 in. and is approxi-mately thick (see the cross-section of the ball in Figure 9.44). Approximately how many cubic inches of plastic were needed to construct the ball? SOLUTION The volume of plastic used is the difference between the outside volume and the inside volume. Where R denotes the length of the outside radius and r denotes the length of the inside radius, and . Then The volume of plastic used was approximately . Like circles, spheres may have tangent lines and secant lines as illustrated in Figure 9.45(a). However, spheres also have tangent planes as shown in Figure 9.45(b).

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  Mathematical Practices, Mathematics for Teachers

  Activities, Models, and Real-Life Examples

  • Ron Larson, Robyn Silbey(Authors)
  • 2014(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Section 12.3 Volumes of Circular Solids 483 Finding the Volume of a Sphere Volume of a Sphere Words The volume V of a sphere is four-thirds Radius, r the product of π and the cube of the radius of the sphere. Algebra V = 4 — 3 πr 3 Cube of radius of sphere EXAMPLE 8 Comparing Volumes of Solids Compare the volumes of the three circular solids. Note that the “height” of the sphere is also 2r. h = 2r r h = 2r r r SOLUTION Cylinder V = Bh = πr 2 h = πr 2 (2r) = 2πr 3 Cone V = 1 — 3 Bh = 1 — 3 πr 2 h = 1 — 3 πr 2 (2r) = 2 — 3 πr 3 ( 1 — 3 ) (2πr 3 ) Sphere V = 4 — 3 πr 3 ( 2 — 3 ) (2π r 3 ) The volume of the cone is 1 — 3 the volume of the cylinder, and the volume of the sphere is 2 — 3 the volume of the cylinder. EXAMPLE 9 Finding the Volume of a Composite Solid The ice cream cone is full of ice cream. Including the scoop on top, find the total volume of ice cream in the cone. Assume that the scoop on top of the cone forms a hemisphere, which is one-half of a sphere. SOLUTION The solid consists of a cone and a hemisphere. You can find the volume of the entire solid by finding the volume of each of these two solids. Volume of cone V = 1 — 3 πr 2 h = 1 — 3 π(1.5) 2 (4) = 3π Volume of hemisphere V = 1 — 2 ⋅ 4 — 3 πr 3 = 1 — 2 ⋅ 4 — 3 π(1.5) 3 = 2.25π So, the volume of ice cream in the cone, including the scoop on top, is 3π + 2.25π = 5.25π, or about 16.5 cubic inches. The point of Example 8 is that the volumes of the three circular solids having the same radius and height are related. The ratio of volumes (cylinder : cone : sphere) is 1 : 1 — 3 : 2 — 3 or 3 : 1 : 2.

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  Geometry

  A Self-Teaching Guide

  • Steve Slavin, Ginny Crisonino(Authors)
  • 2004(Publication Date)
  • Wiley

    (Publisher)

  Volume of a hemisphere formula V = 1 2 4 3 πr 3 , which is 2 3 πr 3 , so V = 2 3 πr 3 Example 16: Find the volume of a hemisphere with a radius of 10 inches. Solution: V = πr 3 = (0.67)(3.14)(10) 3 = (2.1038)(1,000) = 2,103.8 cubic inches Example 17: Find the volume of a sphere with a diameter of 30 inches. Solution: The formula for the volume of a sphere uses the radius, not the diameter, of the sphere. We’ll find the radius by dividing the diameter by 2. r = = = 15 Volume = πr 3 = (1.33)(3.14)(15) 3 = (4.1762)(3,375) = 14,094.7 cubic inches 4 3 30 2 d 2 2 3 4 3 sphere hemisphere 178 GEOMETRY Example 18: Find the volume of a hemisphere with a radius of 4 feet. Solution: V = πr 3 = (.67)(3.14)(4) 3 = (2.1038)(64) = 134.643 cubic feet We’ve looked at the surface area of other three-dimensional solids, so why not a sphere? Surface area of a sphere formula SA = 4πr 2 Example 19: Find the surface area of a sphere with a radius of 5 yards. Solution: SA = 4πr 2 = 4(3.14)(5) 2 = 4(3.14)(25) = 314 square yards Example 20: Find the surface area of a sphere with a diameter of 12.2 yards. Solution: First we have to find the radius of the sphere. r = 2 d = 12 2 .2 = 6.1 SA = 4πr 2 = 4(3.14)(6.1) 2 = (12.56)(37.21) = 467.358 square yards Can you figure out the formula for the surface area of a hemisphere? (Don’t worry about the flat part; just get the spherical part.) Surface area of a hemisphere formula SA = (4πr 2 ) = 2πr 2 , or SA = 2πr 2 Example 21: Find the surface area of a hemisphere with a radius of 2.5 feet. Solution: SA = 2πr 2 = 2(3.14)(2.5) 2 = (6.28)(6.25) = 39.25 square feet 1 2 2 3 Volume and Surface Area of Three-dimensional Polygons 179 SELF-TEST 3 1. Find the volume for a sphere with a radius of a. 2 feet b. 9 inches c. 3.5 yards 2. Find the volume for a sphere with a diameter of 15 inches. 3. Find the volume of a hemisphere with a radius of 18 feet. 4. Find the surface area of a sphere with a radius of a.

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  Calculus: Early Transcendentals, Metric Edition

  • James Stewart, Daniel K. Clegg, Saleem Watson, , James Stewart, James Stewart, Daniel K. Clegg, Saleem Watson(Authors)
  • 2020(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Definition of Volume Let S be a solid that lies between x - a and x - b. If the cross-sectional area of S in the plane P x , through x and perpendicular to the x-axis, is As xd, where A is a continuous function, then the volume of S is V - lim n l` o n i-1 As x i * d Dx - y b a As xd dx It can be proved that this definition is independent of how S is situated with respect to the x-axis. In other words, no matter how we slice S with parallel planes, we always get the same answer for V. When we use the volume formula V - y b a As xd dx , it is important to remember that As xd is the area of a moving cross-section obtained by slicing through x perpendicular to the x-axis. Notice that, for a cylinder, the cross-sectional area is constant: As xd - A for all x. So our definition of volume gives V - y b a A dx - Asb 2 ad; this agrees with the formula V - Ah. EXAMPLE 1 Show that the volume of a sphere of radius r is V - 4 3 r 3 . SOLUTION If we place the sphere so that its center is at the origin, then the plane P x intersects the sphere in a circle whose radius (from the Pythagorean Theorem) is FIGURE 3 Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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  Engineering Mathematics

  • John Bird(Author)
  • 2017(Publication Date)
  • Routledge

    (Publisher)

  Chapter 20 Volumes and surface areas of common solids Why it is important to understand: Volumes and surface areas of common solids There are many practical applications where volumes and surface areas of common solids are required. Examples include determining capacities of oil, water, petrol and fish tanks, ventilation shafts and cooling towers, determining volumes of blocks of metal, ball-bearings, boilers and buoys, and calculating the cubic metres of concrete needed for a path. Finding the surface areas of loudspeaker diaphragms and lampshades provide further practical examples. Understanding these calculations is essential for the many practical applications in engineering, construction, architecture and science. At the end of this chapter, you should be able to: • state the SI unit of volume • calculate the volumes and surface areas of cuboids, cylinders, prisms, pyramids, cones and spheres • calculate volumes and surface areas of frusta of pyramids and cones • calculate the frustum and zone of a sphere • calculate volumes of regular solids using the prismoidal rule • appreciate that volumes of similar bodies are proportional to the cubes of the corresponding linear dimensions 20.1 Introduction The volume of any solid is a measure of the space occupied by the solid. Volume is measured in cubic units such as mm 3 , cm 3 and m 3 . This chapter deals with finding volumes of common solids; in engineering it is often important to be able to calculate volume or capacity to estimate, say, the amount of liquid, such as water, oil or petrol, in differing shaped containers. A prism is a solid with a constant cross-section and with two ends parallel. The shape of the end is used to describe the prism. For example, there are rectangular prisms (called cuboids), triangular prisms and circular prisms (called cylinders).

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  Spherical Geometry and Its Applications

  • Marshall Whittlesey(Author)
  • 2019(Publication Date)
  • CRC Press

    (Publisher)

  Chapter 2

  The sphere in space

  We begin our study of spherical geometry by noting two different ways of presenting the subject, both of which we consider in this book.

  We may first consider a sphere the way most people understand it: as a round object that lives in three-dimensional space. By doing so, we may use the tools of three-dimensional geometry, trigonometry, and vectors to learn about the sphere. This we will call an extrinsic approach to spherical geometry. An understanding of it is particularly important in applications. On the other hand, we may study spherical geometry via the intrinsic properties of the sphere — that is, properties of the sphere that can be thought of without reference to the larger three-dimensional space in which a sphere sits. It will be our main goal to do this in Chapter 3 .

  We begin by looking at the extrinsic properties of the sphere in this chapter. As we do so, it will be important to understand how to see these properties as intrinsic, and hence motivate the axiom system that we set up in Chapter 3 .

  5 Great circles

  Definition 5.1 Let O be a point in space and r a positive real number. The sphere with center O and radius r is the set of all points in space which lie at distance r from O.

  The following proposition summarizes the properties of intersections between spheres and planes in space.

  Proposition 5.2 The intersection between a sphere and a plane in space is a circle, a point, or the empty set.

  In Figure 2.1 we may see these possibilities geometrically by considering the line in space through the center O of the sphere perpendicular to the plane at point P. We distinguish four cases: P is outside the sphere, on the sphere, interior to the sphere but different from O, or equal to O. In the first case, every point in the plane is outside the sphere because P = P1 is the closest point on the plane to O — hence every point on the plane is at a greater distance from O than the radius. If P = P2 is on the sphere, then the plane is tangent to the sphere, hence meets it in a single point. In the case of P = O, the intersection of the sphere with the plane is the set of all points in the plane at a fixed distance from O. This is a circle. Lastly, suppose P = P3 is inside the sphere, but different from O. Then let X be any point on the intersection of the plane and the sphere, let r be the radius of the sphere, and let d be the distance between P3 and O. Then ΔOP3 X has a right angle at P3 , so the length of

  P 3

  X

  ¯

  must be

  r 2

  −

  d 2

  . Thus the points on both the plane and sphere all lie at the same distance from P3 , so lie on a circle. To be complete, we must show that all points in the plane at distance

  r 2

  −

  d 2

  from P3 are also on the sphere; this is left as an exercise. So the intersection of the plane and the sphere is a circle of radius

  r 2

  −

  d 2

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  Science and Mathematics for Engineering

  • John Bird(Author)
  • 2019(Publication Date)
  • Routledge

    (Publisher)

  Chapter 14 Volumes of common solids

  Why it is important to understand: Volumes of common solids

  There are many practical applications where volumes and surface areas of common solids are required. Examples include determining capacities of oil, water, petrol and fish tanks, ventilation shafts and cooling towers, determining volumes of blocks of metal, ball-bearings, boilers and buoys, and calculating the cubic metres of concrete needed for a path. Finding the surface areas of loudspeaker diaphragms and lampshades provide further practical examples. Understanding these calculations is essential for the many practical applications in engineering, construction, architecture and science.

  At the end of this chapter, you should be able to:

  • state the SI unit of volume
  • calculate the volumes and surface areas of cuboids, cylinders, prisms, pyramids, cones and spheres
  • appreciate that volumes of similar bodies are proportional to the cubes of the corresponding linear dimensions

  14.1   Introduction

  The volume of any solid is a measure of the space occupied by the solid. Volume is measured in cubic units such as mm3 , cm3 and m3 .

  This chapter deals with finding volumes of common solids; in engineering it is often important to be able to calculate volume or capacity, to estimate, say, the amount of liquid, such as water, oil or petrol, in differently shaped containers.

  A prism is a solid with a constant cross-section and with two ends parallel. The shape of the end is used to describe the prism. For example, there are rectangular prisms (called cuboids), triangular prisms and circular prisms (called cylinders).

  On completing this chapter you will be able to calculate the volumes and surface areas of rectangular and other prisms, cylinders, pyramids, cones and spheres. Volumes of similar shapes are also considered.

  14.2   Calculating volumes and surface areas of common solids

  14.2.1   Cuboid or rectangular prism

  Science and Mathematics for Engineering. 978-0-367-20475-4, © John Bird. Published by Taylor & Francis. All rights reserved.

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  Solid Geometry with MATLAB Programming

  • Nita H. Shah, Falguni S. Acharya(Authors)
  • 2023(Publication Date)
  • River Publishers

    (Publisher)

  3 Sphere 3.1 Definition The sphere is a locus of a point that moves so that its distance from a fixed point always remains constant. The fixed point is called the center of the sphere and the constant distance is called the radius of the sphere. 3.2 Equation of Sphere in Vector Form Consider a sphere S with center C and radius ‘ a ’. Let P be any point on the sphere. Let c → and r → be the position vectors of C and P respectively with respect to O; the origin of reference. ∴ C P → = O P → − O C → ∴ C P → = r → − c → But. radius C P = | C P → | = a. (3.1) ∴ | r → − c → | = a (3.2) ∴ | r → − c → | 2 = a 2 ∴ (r → − c →). (r → − c →) = a 2 ∴ r →. r → − r →. c → − c →. r → + c →. c → = a 2 ∴. | r → | 2 − 2 r →. c → + | c → | 2 = a 2 ∴ | r → | 2 − 2 r →. c → + d = 0 ∴ | r → | 2 − 2 r →. c → = | c → | 2 = a 2 where d = | c → | 2 − a 2 is the equation of sphere in vector form. Case 1): If O lies on the sphere, then OC = OP; i.e., c. = a ⇒ d = 0 The Equation of sphere is | r → | 2 − 2 r →. c → = 0 Case 2): If the center C is at the origin O, then c = 0 and d = − a 2. ∴ Equation of sphere is | r → | 2 − 2 r →. c → = 0 | r → | 2 = a 2 | r → | = a. Equation of sphere in the cartesian form: Consider a sphere S with center C (α, β, γ), and radius ‘ a ’. Let P (x, y, z) be any point on the sphere. Let x → and c → be the position vector of P and C respectively with respect to O; the origin of reference. ∴ r → = x i → + y j → + z k → c → = α i → + β j → + γ k →. The equation in vector form. is | r → − c → | = a. ∴ | (x − α) i → + (y − β) j → + (x − γ) k → | = a ∴ (x − α) 2 + (y − β) 2 + (z − γ) 2 = a 2 which is the equation of sphere S in cartesian form with center C (α, β, γ), and radius ‘ a ’. Remark: When center C is origin. i.e., α = β = γ = 0 then the equation of the sphere is x 2 + y 2 + z 2 = a 2. 3.3 General Equation of the Sphere The general equation of the sphere

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