Learn on PengiReveal Math, Course 3Module 10: Volume

Lesson 10-5: Volume of Composite Solids

Property.

Section 1

Volume Formulas for Rectangular Prisms and Cubes

Property

Cube whose edge is of length LL:

Volume =L3= L^3

Section 2

Volume Of A Cylinder

Property

A cylinder's volume is found using V=BhV = Bh. Since the base 'B' is a circle with area B=πr2B = \pi r^2, the specific formula for a cylinder becomes V=πr2hV = \pi r^2 h.

Examples

A can with radius 3 cm and height 10 cm has a volume of V=π(3 cm)2(10 cm)=90π cm3V = \pi (3 \text{ cm})^2(10 \text{ cm}) = 90\pi \text{ cm}^3.
A battery with a diameter of 14 mm (radius 7 mm) and height 50 mm has a volume of V227(7 mm)2(50 mm)=7700 mm3V \approx \frac{22}{7} (7 \text{ mm})^2(50 \text{ mm}) = 7700 \text{ mm}^3.

Explanation

Think of a can of Pringles. First, find the area of a single chip (the base 'B') using πr2\pi r^2. Then, just multiply that base area by the height of the can 'h' to find its total volume. Remember to use the radius, which is half the diameter, to get the right answer!

Section 3

Volume of a Cone

Property

The volume VV of a cone with radius rr and height hh is given by the formula:

V=13πr2hV = \frac{1}{3}\pi r^2 h

Examples

  • A cone with a radius of 33 cm and a height of 55 cm has a volume of V=13π(32)(5)=15πV = \frac{1}{3}\pi (3^2)(5) = 15\pi cm3^3.
  • A cone with a radius of 44 inches and a height of 99 inches has a volume of V=13π(42)(9)=48πV = \frac{1}{3}\pi (4^2)(9) = 48\pi inches3^3.

Explanation

The volume of a cone measures the amount of space it occupies. This formula shows that the volume depends on the radius of its circular base (rr) and its perpendicular height (hh). An important relationship to note is that a cone''s volume is exactly one-third the volume of a cylinder with the same radius and height. To calculate the volume, substitute the known values for the radius and height into the formula.

Section 4

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}.

Book overview

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

Continue this chapter

Module 10: Volume

  1. Lesson 1

    Lesson 10-1: Volume of Cylinders

  2. Lesson 2

    Lesson 10-2: Volume of Cones

  3. Lesson 3

    Lesson 10-3: Volume of Spheres

  4. Lesson 4

    Lesson 10-4: Find Missing Dimensions

  5. Lesson 5Current

    Lesson 10-5: Volume of Composite Solids

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Volume Formulas for Rectangular Prisms and Cubes

Property

Cube whose edge is of length LL:

Volume =L3= L^3

Section 2

Volume Of A Cylinder

Property

A cylinder's volume is found using V=BhV = Bh. Since the base 'B' is a circle with area B=πr2B = \pi r^2, the specific formula for a cylinder becomes V=πr2hV = \pi r^2 h.

Examples

A can with radius 3 cm and height 10 cm has a volume of V=π(3 cm)2(10 cm)=90π cm3V = \pi (3 \text{ cm})^2(10 \text{ cm}) = 90\pi \text{ cm}^3.
A battery with a diameter of 14 mm (radius 7 mm) and height 50 mm has a volume of V227(7 mm)2(50 mm)=7700 mm3V \approx \frac{22}{7} (7 \text{ mm})^2(50 \text{ mm}) = 7700 \text{ mm}^3.

Explanation

Think of a can of Pringles. First, find the area of a single chip (the base 'B') using πr2\pi r^2. Then, just multiply that base area by the height of the can 'h' to find its total volume. Remember to use the radius, which is half the diameter, to get the right answer!

Section 3

Volume of a Cone

Property

The volume VV of a cone with radius rr and height hh is given by the formula:

V=13πr2hV = \frac{1}{3}\pi r^2 h

Examples

  • A cone with a radius of 33 cm and a height of 55 cm has a volume of V=13π(32)(5)=15πV = \frac{1}{3}\pi (3^2)(5) = 15\pi cm3^3.
  • A cone with a radius of 44 inches and a height of 99 inches has a volume of V=13π(42)(9)=48πV = \frac{1}{3}\pi (4^2)(9) = 48\pi inches3^3.

Explanation

The volume of a cone measures the amount of space it occupies. This formula shows that the volume depends on the radius of its circular base (rr) and its perpendicular height (hh). An important relationship to note is that a cone''s volume is exactly one-third the volume of a cylinder with the same radius and height. To calculate the volume, substitute the known values for the radius and height into the formula.

Section 4

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}.

Book overview

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

Continue this chapter

Module 10: Volume

  1. Lesson 1

    Lesson 10-1: Volume of Cylinders

  2. Lesson 2

    Lesson 10-2: Volume of Cones

  3. Lesson 3

    Lesson 10-3: Volume of Spheres

  4. Lesson 4

    Lesson 10-4: Find Missing Dimensions

  5. Lesson 5Current

    Lesson 10-5: Volume of Composite Solids