Learn on PengiBig Ideas Math, Course 3Chapter 8: Volume and Similar Solids

Lesson 3: Volumes of Spheres

In this Grade 8 lesson from Big Ideas Math Course 3, Chapter 8, students learn how to calculate the volume of a sphere using the formula V = (4/3)πr³, which is derived by comparing a sphere's volume to that of a cylinder and through a pyramid-based approach. Students practice applying the formula to find volume given a radius and to find the radius when volume is known, working with concepts such as hemisphere and surface area of a sphere.

Section 1

The Sphere

Property

A sphere of radius rr is the set of all points in space that are of a distance rr from a point CC, called the center of the sphere. The volume of a sphere of radius rr is V=43πr3V = \frac{4}{3} \pi r^3.

Examples

  • A basketball has a radius of 12 cm. Its volume is V=43π(123)=43π(1728)=2304πV = \frac{4}{3} \pi (12^3) = \frac{4}{3} \pi (1728) = 2304\pi cubic cm.
  • A hemisphere is half of a sphere. If a bowl in the shape of a hemisphere has a radius of 5 inches, its volume is V=1243π(53)=23π(125)=2503πV = \frac{1}{2} \cdot \frac{4}{3} \pi (5^3) = \frac{2}{3} \pi (125) = \frac{250}{3}\pi cubic inches.
  • A spherical balloon has a volume of 36π36\pi cubic feet. To find its radius, we solve 36π=43πr336\pi = \frac{4}{3} \pi r^3. This simplifies to 27=r327 = r^3, so the radius is 3 feet.

Explanation

This formula calculates the space inside a perfectly round ball. The only measurement you need is the radius, which is the distance from the center to the surface. Notice the radius is cubed (r3r^3), reflecting its three-dimensional nature.

Section 2

Sphere Volume from Cylinder Relationship

Property

The volume of a sphere is 23\frac{2}{3} times the volume of a cylinder that has the same diameter as the sphere and height equal to the diameter:

Vsphere=23×Vcylinder=23×πr2hV_{sphere} = \frac{2}{3} \times V_{cylinder} = \frac{2}{3} \times \pi r^2 h

When h=2rh = 2r (diameter), this becomes:

Vsphere=23×πr2(2r)=43πr3V_{sphere} = \frac{2}{3} \times \pi r^2 (2r) = \frac{4}{3}\pi r^3

Section 3

Finding Radius from Volume

Property

To find the radius of a sphere when given its volume, solve the equation V=43πr3V = \frac{4}{3}\pi r^3 for rr:

r=3V4π3r = \sqrt[3]{\frac{3V}{4\pi}}

Book overview

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Chapter 8: Volume and Similar Solids

  1. Lesson 1

    Lesson 1: Volumes of Cylinders

  2. Lesson 2

    Lesson 2: Volumes of Cones

  3. Lesson 3Current

    Lesson 3: Volumes of Spheres

  4. Lesson 4

    Lesson 4: Surface Areas and Volumes of Similar Solids

Lesson overview

Expand to review the lesson summary and core properties.

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Section 1

The Sphere

Property

A sphere of radius rr is the set of all points in space that are of a distance rr from a point CC, called the center of the sphere. The volume of a sphere of radius rr is V=43πr3V = \frac{4}{3} \pi r^3.

Examples

  • A basketball has a radius of 12 cm. Its volume is V=43π(123)=43π(1728)=2304πV = \frac{4}{3} \pi (12^3) = \frac{4}{3} \pi (1728) = 2304\pi cubic cm.
  • A hemisphere is half of a sphere. If a bowl in the shape of a hemisphere has a radius of 5 inches, its volume is V=1243π(53)=23π(125)=2503πV = \frac{1}{2} \cdot \frac{4}{3} \pi (5^3) = \frac{2}{3} \pi (125) = \frac{250}{3}\pi cubic inches.
  • A spherical balloon has a volume of 36π36\pi cubic feet. To find its radius, we solve 36π=43πr336\pi = \frac{4}{3} \pi r^3. This simplifies to 27=r327 = r^3, so the radius is 3 feet.

Explanation

This formula calculates the space inside a perfectly round ball. The only measurement you need is the radius, which is the distance from the center to the surface. Notice the radius is cubed (r3r^3), reflecting its three-dimensional nature.

Section 2

Sphere Volume from Cylinder Relationship

Property

The volume of a sphere is 23\frac{2}{3} times the volume of a cylinder that has the same diameter as the sphere and height equal to the diameter:

Vsphere=23×Vcylinder=23×πr2hV_{sphere} = \frac{2}{3} \times V_{cylinder} = \frac{2}{3} \times \pi r^2 h

When h=2rh = 2r (diameter), this becomes:

Vsphere=23×πr2(2r)=43πr3V_{sphere} = \frac{2}{3} \times \pi r^2 (2r) = \frac{4}{3}\pi r^3

Section 3

Finding Radius from Volume

Property

To find the radius of a sphere when given its volume, solve the equation V=43πr3V = \frac{4}{3}\pi r^3 for rr:

r=3V4π3r = \sqrt[3]{\frac{3V}{4\pi}}

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 8: Volume and Similar Solids

  1. Lesson 1

    Lesson 1: Volumes of Cylinders

  2. Lesson 2

    Lesson 2: Volumes of Cones

  3. Lesson 3Current

    Lesson 3: Volumes of Spheres

  4. Lesson 4

    Lesson 4: Surface Areas and Volumes of Similar Solids