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Curved-Surface Solids

Curved-surface solids include cylinders, cones, and spheres. This article focuses on the cylinder, a solid formed by two parallel congruent circles and one curved lateral surface.

3.7 Generalize the surface area and volume of curved-surface solids such as cylinders, cones, and spheres.
4.7 Solve contextual problems related to the surface area and volume of curved-surface solids and their combinations.

After studying this lesson, students are expected to be able to:

  1. identify the elements of a cylinder
  2. calculate the base area, lateral area, and top area
  3. calculate the volume of a cylinder
  4. determine cylinder elements when the volume is known
  5. compare cylinder volumes after a change in radius
  6. calculate the change in cylinder volume when the radius changes

A cylinder is a curved-surface solid formed by two identical parallel circles and a rectangle wrapped around them.

Calculating the area and volume of a cylinder

From the picture we see:

  1. a cylinder has three surfaces: the base, the top, and the curved side
  2. the base and top are circles
  3. the curved side is called the lateral surface
  4. the distance between base and top is the height, denoted by tt
  5. the radius is denoted by rr and the diameter by dd

A cylinder net consists of:

  • two congruent circles
  • one rectangle

Cylinder net

The net is useful because the surface area of the cylinder is equal to the total area of this unfolded figure.

The surface area is the sum of:

  1. the area of the base circle
  2. the area of the lateral surface
  3. the area of the top circle

Details:

  • base area = πr2\pi r^2
  • lateral area = circumference of base ×\times height = 2πrt2\pi rt
  • top area = πr2\pi r^2

So:

S=πr2+2πrt+πr2=2πr2+2πrt=2πr(r+t) \begin{aligned} S &= \pi r^2 + 2\pi rt + \pi r^2\\ &= 2\pi r^2 + 2\pi rt\\ &= 2\pi r(r+t) \end{aligned}

A cylinder has height 13 cm and base radius 7 cm. Find its surface area.

S=2πr(r+t)=2×227×7×(7+13)=44×20=880 \begin{aligned} S &= 2\pi r(r+t)\\ &= 2 \times \frac{22}{7} \times 7 \times (7+13)\\ &= 44 \times 20\\ &= 880 \end{aligned}

Therefore:

880 cm2 880\ \text{cm}^2

Since the base is a circle, the volume of a cylinder is the base area times the height.

Formula:

V=πr2t V = \pi r^2 t

or, in terms of diameter:

V=14πd2t V = \frac{1}{4}\pi d^2 t

A cylinder has radius 14 cm and height 20 cm. Find its volume.

V=πr2t=227×142×20=12,320 \begin{aligned} V &= \pi r^2 t\\ &= \frac{22}{7}\times 14^2 \times 20\\ &= 12{,}320 \end{aligned}

So the volume is:

12,320 cm3 12{,}320\ \text{cm}^3

A cylindrical drinking container has volume 693 ml. If its height is 18 cm and the container is full, determine:

  1. the diameter
  2. the total surface area

Since 1 ml = 1 cm^3, we have:

V=693 cm3 V = 693\ \text{cm}^3 V=πr2t693=3.14×r2×18693=56.52r2r2=69356.52r212.26r3.5 \begin{aligned} V &= \pi r^2 t\\ 693 &= 3.14 \times r^2 \times 18\\ 693 &= 56.52r^2\\ r^2 &= \frac{693}{56.52}\\ r^2 &\approx 12.26\\ r &\approx 3.5 \end{aligned}

So:

d=2r=7 cm d = 2r = 7\ \text{cm} S=2πr(r+t)=2×227×3.5×(3.5+18)=473 \begin{aligned} S &= 2\pi r(r+t)\\ &= 2 \times \frac{22}{7} \times 3.5 \times (3.5+18)\\ &= 473 \end{aligned}

Therefore:

473 cm2 473\ \text{cm}^2
  • base area = πr2\pi r^2
  • lateral area = 2πrt2\pi rt
  • top area = πr2\pi r^2
  • total surface area = 2πr(r+t)2\pi r(r+t)
  • surface area without top = πr(r+2t)\pi r(r+2t)
  • volume = πr2t\pi r^2t

Use the following simulation to explore how changes in radius and height affect the surface area and volume of a cylinder.

Try the following problems:

  1. A cylinder has height 14 cm and base radius 3 cm. Find its volume.
  2. A cylinder is made of sheet metal with radius 14 cm and height 20 cm. Find the sheet area needed to make it.
  3. A cylindrical water tank has height 2 m and diameter 7 dm. The bottom leaks and water flows out at an average rate of 5 liters per minute. If the tank is full, after how many minutes will it be empty?

The cylinder is one of the most common curved-surface solids in daily life, appearing in cans, pipes, glasses, and water tanks. Understanding its elements, surface area, and volume makes many applied geometry problems much easier to solve.

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