(*) Volume of Cylinders

Subsection 2.8.3 (*) Volume of Cylinders

Curriculum Alignment

Strand 🔗: 2.0 Measurements and Geometry
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Sub-Strand 🔗: 2.8 Volume
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Determine the volume of composite solids

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Explore the use of the surface area and volume of solids in real-life situations

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Learner Experience 2.8.3.

Work in groups:

Materials Needed: A cylinder-shaped container (e.g., a can or a water bottle), a measuring tape, water, and a measuring cup.

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Instruction:

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Form groups of $2-3$ students.

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Find the radius and height of the cylinder-shaped container (i.e. a can) using the measuring tape, in centimeters.
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Pour water to fill the cylinder.
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Transfer the water from the cylinder to the measuring cup and record the volume of water in milliliters (mL).
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Calculate the volume of the cylinder using the formula $V = \pi r^2 h\text{,}$ where $r$ is the radius and $h$ is the height. Use $\pi = 3.14\text{.}$
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Compare your calculated volume with milliliters you measured (mL). Discuss any differences and possible reasons for them.
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Key Takeaway 2.8.14.

The volume of a cylinder can be calculated using the formula

\begin{equation*} V = \pi r^2 h\text{,} \end{equation*}

where $r$ is the radius of the circular base and $h$ is the height of the cylinder.

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This formula is just a special case of the volume formula for prisms ($V = Fh$), since the area of a circular base is given by $F = \pi r^2\text{.}$

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Example 2.8.15.

The cylinder below has a radius of 4 cm and a height of 10 cm. Use $\pi = 3.142$

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$Cylindrical prism$

\begin{align*} V \amp = \text{Base Area }\times h \\ V \amp = \pi r^2 \times h\\ \amp = 3.142 \times (4)^2 \, \text{cm} \times 10\, \text{cm}\\ \amp = 502.65\, \text{cm}^3 \end{align*}

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Example 2.8.16.

A pipe has a radius of 3 cm and a height of 15 cm. Find the volume of the pipe.

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Solution.

\begin{align*} V \amp = \text{Base Area} \times h \\ V \amp = \pi r^2 \times h\\ \amp = 3.142 \times (3)^2 \, \text{cm} \times 15\, \text{cm}\\ \amp = 424.12\, \text{cm}^3 \end{align*}

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Example 2.8.17.

Find the volume of the following cylinder using $\pi = 3.142\text{.}$ Leave your answer (correct to 2 decimal place):

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Solution.

Step 1: Find the area of the base

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\begin{align*} \text {area of circle } = \amp \pi r^2 \\ = \amp \pi(8)^2\\ = \amp \pi 64\,\text{cm}^2\\ = \amp 3,142 \times 64\,\text{cm}^2 \\ = \amp 201.088 \,\text{cm}^2 \end{align*}

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Step 2: Multiply the area of the base by the height of the solid to find the volume

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\begin{align*} \text{Volume} = \amp \text{Base area} \times \text{height}\\ = \amp \pi r^2 \times h \\ = \amp201.088 \, \text{cm}^2 \times 20\, \text{cm} \\ = \amp 4021.76 \, \text{cm}^3 \end{align*}

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The volume of the cylinder is $4021.76 \, \text{cm}^3 $

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Example 2.8.18.

Mueni’s water tank has a radius of 7 m and a height of 20 m. Find the capacity of the water tank in litres.

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Solution.

\begin{align*} V \amp = \text{Base Area} \times h \\ V \amp = \pi r^2 \times h\\ \amp = \frac{22}{7} \times (7)^2\, \text{m} \times 20 \, \text{m}\\ \amp = 3080\, \text{m}^3 \end{align*}

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To convert from $m^3 $ to litres we multiply by $1000 \, L/\text{m}^3$

\begin{align*} 1 \text{litre} \amp = 1000\, \text{m}^3 \\ \amp = 3080\, \text{m}^3 \times 1000\, \text{L/m}^3\\ \amp = 3,080,000 \, \text{ litres} \end{align*}

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Checkpoint 2.8.19.

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Checkpoint 2.8.20.

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Exercises Exercises

Unless otherwise stated, use $\pi = 3.14$ in all calculations.

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1.

Alongside is a cross-sectional view of an open cylinder that has a radius of 7 cm and a height of 10 cm. Find its volume.

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Solution.

Use $V=\pi r^2 h$ with $r=7\,\text{cm}$ and $h=10\,\text{cm}\text{.}$

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\begin{align*} V \amp = \pi \times 7^2 \times 10 \\ \amp = \pi \times 49 \times 10 \\ \amp = 490\pi\,\text{cm}^3 \\ \amp \approx 3.14 \times 490 = 1538.6\,\text{cm}^3 \end{align*}

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2.

Calculate the volume of a cylinder with radius $6\,\text{cm}$ and height $10\,\text{cm}\text{.}$ (Use $\pi=3.14\text{.}$)

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Solution.

Using $V=\pi r^2 h$ we have

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\begin{align*} V\amp=3.14\times 6^2 \times 10\\ \amp=3.14\times 36 \times 10\\ \amp=1130.4\,\text{cm}^3 \end{align*}

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3.

A cylinder has diameter $10\,\text{cm}$ and height $12\,\text{cm}\text{.}$ Find its volume in terms of $\pi\text{.}$

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Solution.

The radius is $r=\tfrac{10}{2}=5\,\text{cm}\text{.}$ Using $V=\pi r^2 h\text{:}$

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\begin{align*} V\amp=\pi\times 5^2 \times 12\\ \amp=\pi \times 25 \times 12\\ \amp=300\pi\,\text{cm}^3 \end{align*}

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4.

A cylinder has a volume of $1,570 \text{cm}^3$ and a diameter of 30 cm. Find its height.

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Solution.

Radius $r=\tfrac{30}{2}=15\,\text{cm}\text{.}$ Using $V=\pi r^2 h$ we get

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\begin{align*} h\amp=\dfrac{V}{\pi r^2} = \dfrac{1570}{3.14\times 15^2}\\ \amp=\dfrac{1570}{706.5}\\ \amp\approx 2.22\,\text{cm} \end{align*}

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5.

A cylindrical water tank has a radius of 2 m and a height of 5 m. How many liters of water can it hold when full? (Use $\pi = 3.14\text{,}$ remember that 1 cubic meter = 1000 liters)

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Solution.

Using $V=\pi r^2 h$ we have

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\begin{align*} V \amp = \pi r^2 h \\ \amp = 3.14 \times (2)^2\, \text{m} \times 5\, \text{m} \\ \amp = 3.14 \times 4 \times 5\, \text{m}^3 \\ \amp = 62.8\, \text{m}^3 \end{align*}

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To convert from $\text{m}^3$ to litres we multiply by $1000\,\text{L/m}^3\text{.}$

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\begin{align*} V = 62.8\, \text{m}^3 \times 1000\, \text{L/m}^3 \\ \amp = 62,800\, \text{litres} \end{align*}

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6.

If the radius of a cylinder is doubled while keeping the height the same, by what factor does the volume increase?

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Solution.

Let the original radius be $r$ and height $h\text{.}$ Original volume: $V=\pi r^2 h\text{.}$

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If radius is doubled ($r'=2r$) then

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\begin{align*} V' \amp = \pi (2r)^2 h = \pi \times 4r^2 h = 4\pi r^2 h \\ \amp = 4V \end{align*}

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Therefore the volume increases by a factor of $4\text{.}$

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7. A Harder Exercise.

A hollow cylindrical pipe has an outer radius of 8 cm, an inner radius of 6 cm, and a length of 100 cm. Find the volume of material used to make the pipe.

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Solution.

The volume of material is the volume of the outer cylinder minus the inner cylinder: $V=\pi(R^2-r^2)h\text{.}$

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Here $R=8\,\text{cm}\text{,}$ $r=6\,\text{cm}$ and $h=100\,\text{cm}\text{.}$ So

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\begin{align*} V \amp = \pi(8^2-6^2)\times 100\\ \amp = \pi(64-36)\times 100\\ \amp = \pi\times 28 \times 100\\ \amp = 2800\pi\,\text{cm}^3 \end{align*}

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Using $\pi=3.14$ (unless otherwise stated) gives

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\begin{align*} V \amp \approx 3.14 \times 2800 = 8792\,\text{cm}^3 \end{align*}

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