Learn on PengiBig Ideas Math, Advanced 2Chapter 14: Surface Area and Volume

Section 14.3: Surface Areas of Cylinders

In this Grade 7 lesson from Big Ideas Math Advanced 2, Chapter 14, students learn how to find the surface area of a cylinder using the formula S = 2πr² + 2πrh, which accounts for both the two circular bases and the lateral surface. Students also practice finding lateral surface area alone, as when calculating the area of a label on a can. Real-world applications include estimating surface areas of everyday cylindrical containers and using proportional reasoning to solve problems.

Section 1

Cylinder Structure and Components

Property

A cylinder is a three-dimensional solid with two parallel circular bases connected by a curved surface. The cylinder has three key components for surface area calculations: two circular bases each with area πr2\pi r^2, and a rectangular lateral (side) surface that wraps around the cylinder. When "unrolled," this lateral surface forms a rectangle with width equal to the circumference of the base (2πr2\pi r) and height equal to the cylinder's height (hh).

Examples

  • A soup can with radius 3 cm and height 10 cm has two circular bases each with area π(32)=9π\pi(3^2) = 9\pi square cm, and a lateral surface that unrolls into a rectangle with dimensions 2π(3)=6π2\pi(3) = 6\pi cm by 10 cm.
  • A cylindrical water tank with radius 5 feet and height 8 feet consists of two circular ends each with area 25π25\pi square feet, plus a curved side surface with circumference 10π10\pi feet wrapping around the 8-foot height.
  • A paper towel tube with radius 2 inches and height 12 inches has circular ends with area 4π4\pi square inches each, and if you cut and unroll the cardboard tube, it forms a rectangle that is 4π4\pi inches wide and 12 inches tall.

Explanation

Understanding cylinder structure is essential for calculating surface area. Think of a cylinder as being made from two circular "caps" plus a rectangular piece of material wrapped around to form the sides. The key insight is that the curved lateral surface, when flattened out, becomes a rectangle whose width matches exactly the circumference of the circular base.

Section 2

Surface Area of Cylinders

Property

Cylinder with radius rr and height hh:
Surface Area =2πr2+2πrh= 2\pi r^2 + 2\pi rh
This consists of:

  • Two circular bases: 2πr22\pi r^2
  • Curved lateral surface: 2πrh2\pi rh

Examples

  • A cylinder with radius 3 cm and height 8 cm has a Surface Area of 2π(32)+2π(3)(8)=2π(9)+2π(24)=18π+48π=66π2\pi(3^2) + 2\pi(3)(8) = 2\pi(9) + 2\pi(24) = 18\pi + 48\pi = 66\pi square cm, which is approximately 207.3207.3 square cm.
  • A cylindrical can with radius 2 inches and height 5 inches has a Surface Area of 2π(22)+2π(2)(5)=8π+20π=28π2\pi(2^2) + 2\pi(2)(5) = 8\pi + 20\pi = 28\pi square inches, which is approximately 87.987.9 square inches.
  • A cylinder with radius 4 ft and height 6 ft has a Surface Area of 2π(42)+2π(4)(6)=32π+48π=80π2\pi(4^2) + 2\pi(4)(6) = 32\pi + 48\pi = 80\pi square ft, which is approximately 251.3251.3 square ft.

Explanation

The surface area of a cylinder measures the total area of all its outside surfaces. This includes the two circular bases (top and bottom) plus the curved surface that wraps around the side. Think of it as the amount of material needed to completely cover the cylinder, like wrapping paper or paint.

Section 3

Scaling Effects on Cylinder Surface Area

Property

When a cylinder is scaled by a length factor of kk, its surface area is scaled by a factor of k2k^2.

If the original cylinder has surface area SASA, then the scaled cylinder has surface area k2SAk^2 \cdot SA.

Book overview

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Chapter 14: Surface Area and Volume

  1. Lesson 1

    Section 14.1: Surface Areas of Prisms

  2. Lesson 2

    Section 14.2: Surface Areas of Pyramids

  3. Lesson 3Current

    Section 14.3: Surface Areas of Cylinders

  4. Lesson 4

    Section 14.4: Volumes of Prisms

  5. Lesson 5

    Section 14.5: Volumes of Pyramids

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Cylinder Structure and Components

Property

A cylinder is a three-dimensional solid with two parallel circular bases connected by a curved surface. The cylinder has three key components for surface area calculations: two circular bases each with area πr2\pi r^2, and a rectangular lateral (side) surface that wraps around the cylinder. When "unrolled," this lateral surface forms a rectangle with width equal to the circumference of the base (2πr2\pi r) and height equal to the cylinder's height (hh).

Examples

  • A soup can with radius 3 cm and height 10 cm has two circular bases each with area π(32)=9π\pi(3^2) = 9\pi square cm, and a lateral surface that unrolls into a rectangle with dimensions 2π(3)=6π2\pi(3) = 6\pi cm by 10 cm.
  • A cylindrical water tank with radius 5 feet and height 8 feet consists of two circular ends each with area 25π25\pi square feet, plus a curved side surface with circumference 10π10\pi feet wrapping around the 8-foot height.
  • A paper towel tube with radius 2 inches and height 12 inches has circular ends with area 4π4\pi square inches each, and if you cut and unroll the cardboard tube, it forms a rectangle that is 4π4\pi inches wide and 12 inches tall.

Explanation

Understanding cylinder structure is essential for calculating surface area. Think of a cylinder as being made from two circular "caps" plus a rectangular piece of material wrapped around to form the sides. The key insight is that the curved lateral surface, when flattened out, becomes a rectangle whose width matches exactly the circumference of the circular base.

Section 2

Surface Area of Cylinders

Property

Cylinder with radius rr and height hh:
Surface Area =2πr2+2πrh= 2\pi r^2 + 2\pi rh
This consists of:

  • Two circular bases: 2πr22\pi r^2
  • Curved lateral surface: 2πrh2\pi rh

Examples

  • A cylinder with radius 3 cm and height 8 cm has a Surface Area of 2π(32)+2π(3)(8)=2π(9)+2π(24)=18π+48π=66π2\pi(3^2) + 2\pi(3)(8) = 2\pi(9) + 2\pi(24) = 18\pi + 48\pi = 66\pi square cm, which is approximately 207.3207.3 square cm.
  • A cylindrical can with radius 2 inches and height 5 inches has a Surface Area of 2π(22)+2π(2)(5)=8π+20π=28π2\pi(2^2) + 2\pi(2)(5) = 8\pi + 20\pi = 28\pi square inches, which is approximately 87.987.9 square inches.
  • A cylinder with radius 4 ft and height 6 ft has a Surface Area of 2π(42)+2π(4)(6)=32π+48π=80π2\pi(4^2) + 2\pi(4)(6) = 32\pi + 48\pi = 80\pi square ft, which is approximately 251.3251.3 square ft.

Explanation

The surface area of a cylinder measures the total area of all its outside surfaces. This includes the two circular bases (top and bottom) plus the curved surface that wraps around the side. Think of it as the amount of material needed to completely cover the cylinder, like wrapping paper or paint.

Section 3

Scaling Effects on Cylinder Surface Area

Property

When a cylinder is scaled by a length factor of kk, its surface area is scaled by a factor of k2k^2.

If the original cylinder has surface area SASA, then the scaled cylinder has surface area k2SAk^2 \cdot SA.

Book overview

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

Continue this chapter

Chapter 14: Surface Area and Volume

  1. Lesson 1

    Section 14.1: Surface Areas of Prisms

  2. Lesson 2

    Section 14.2: Surface Areas of Pyramids

  3. Lesson 3Current

    Section 14.3: Surface Areas of Cylinders

  4. Lesson 4

    Section 14.4: Volumes of Prisms

  5. Lesson 5

    Section 14.5: Volumes of Pyramids