Learn on PengiPengi Math (Grade 8)Chapter 7: The Pythagorean Theorem and Volume

Lesson 5: Volume of Spheres and Composite Solids

In this Grade 8 lesson from Pengi Math Chapter 7, students learn to apply the sphere volume formula (V = 4/3πr³) and explore how the volumes of spheres, cylinders, and cones relate to one another. Students practice calculating volumes of hemispheres and composite solids such as silos and ice cream cones. The lesson also covers real-world word problems involving density and capacity.

Section 1

Volume of a Sphere

Property

The volume of a sphere is given by

Volume=43×πr3\text{Volume} = \dfrac{4}{3} \times \pi r^3

where rr is the radius of the sphere. Recall that r3r^3, which we read as 'rr cubed,' means r×r×rr \times r \times r.

Examples

  • A gumball has a radius of 1 centimeter. Its volume is V=43π(1)3=43π4.19V = \frac{4}{3} \pi (1)^3 = \frac{4}{3}\pi \approx 4.19 cubic centimeters.
  • A soccer ball has a diameter of 22 cm, so its radius is 11 cm. Its volume is V=43π(11)3=43π(1331)5575.28V = \frac{4}{3} \pi (11)^3 = \frac{4}{3} \pi (1331) \approx 5575.28 cubic centimeters.
  • A spherical ornament has a volume of 36π36\pi cubic inches. To find its radius, solve 36π=43πr336\pi = \frac{4}{3}\pi r^3, which simplifies to 27=r327 = r^3, so the radius is r=273=3r = \sqrt[3]{27} = 3 inches.

Explanation

Volume measures the space inside a 3D shape, like a ball or a planet. For a sphere, you cube the radius (multiply it by itself three times), then multiply by pi (π\pi), and finally multiply by the fraction 43\frac{4}{3}.

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

Volume of Hemispheres

Property

The volume of a hemisphere is half the volume of a complete sphere:

Vhemisphere=1243πr3=23πr3V_{hemisphere} = \frac{1}{2} \cdot \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3

Examples

Book overview

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Chapter 7: The Pythagorean Theorem and Volume

  1. Lesson 1

    Lesson 1: Understanding and Proving the Pythagorean Theorem

  2. Lesson 2

    Lesson 2: Solving Problems Using the Pythagorean Theorem

  3. Lesson 3

    Lesson 3: Distance on the Coordinate Plane

  4. Lesson 4

    Lesson 4: Volume of Cylinders and Cones

  5. Lesson 5Current

    Lesson 5: Volume of Spheres and Composite Solids

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Volume of a Sphere

Property

The volume of a sphere is given by

Volume=43×πr3\text{Volume} = \dfrac{4}{3} \times \pi r^3

where rr is the radius of the sphere. Recall that r3r^3, which we read as 'rr cubed,' means r×r×rr \times r \times r.

Examples

  • A gumball has a radius of 1 centimeter. Its volume is V=43π(1)3=43π4.19V = \frac{4}{3} \pi (1)^3 = \frac{4}{3}\pi \approx 4.19 cubic centimeters.
  • A soccer ball has a diameter of 22 cm, so its radius is 11 cm. Its volume is V=43π(11)3=43π(1331)5575.28V = \frac{4}{3} \pi (11)^3 = \frac{4}{3} \pi (1331) \approx 5575.28 cubic centimeters.
  • A spherical ornament has a volume of 36π36\pi cubic inches. To find its radius, solve 36π=43πr336\pi = \frac{4}{3}\pi r^3, which simplifies to 27=r327 = r^3, so the radius is r=273=3r = \sqrt[3]{27} = 3 inches.

Explanation

Volume measures the space inside a 3D shape, like a ball or a planet. For a sphere, you cube the radius (multiply it by itself three times), then multiply by pi (π\pi), and finally multiply by the fraction 43\frac{4}{3}.

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

Volume of Hemispheres

Property

The volume of a hemisphere is half the volume of a complete sphere:

Vhemisphere=1243πr3=23πr3V_{hemisphere} = \frac{1}{2} \cdot \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3

Examples

Book overview

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

Continue this chapter

Chapter 7: The Pythagorean Theorem and Volume

  1. Lesson 1

    Lesson 1: Understanding and Proving the Pythagorean Theorem

  2. Lesson 2

    Lesson 2: Solving Problems Using the Pythagorean Theorem

  3. Lesson 3

    Lesson 3: Distance on the Coordinate Plane

  4. Lesson 4

    Lesson 4: Volume of Cylinders and Cones

  5. Lesson 5Current

    Lesson 5: Volume of Spheres and Composite Solids