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# 8.6: Cylinder: Base Area, Lateral Area, Surface Area and Volume

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Find the base area, lateral area, and total surface area of cylinders.
• Find the volume of cylinders.

## Cylinders

A cylinder is a three-dimensional figure with a pair of parallel and congruent circular ends, or bases. A cylinder has a single curved side that forms a rectangle when laid out flat.

• A cylinder has bases in the shape of __________________________.
• When flattened, the side of a cylinder is a ______________________________.

As with prisms, cylinders can be right or oblique.

The side of a right cylinder is perpendicular to its circular bases.

The side of an oblique cylinder is not perpendicular to its bases.

• Right cylinders have a side that is ____________________________to its bases. (In this lesson, we will focus on right cylinders only.)

## Surface Area of a Cylinder Using Nets

You can deconstruct a cylinder into a net:

The area of each base is given by the area of a circle:

\begin{align*}A &= \pi r^2\\ &= \pi (5)^2\\ &= 25 \pi \quad \text{or}\\ & \approx (25)(3.14) = 78.5\end{align*}

The area of the rectangular lateral area \begin{align*}L\end{align*} is given by the product of a width and height. The height is given as 24. You can see that the width of the area is equal to the circumference of the circular base:

• The width of the lateral area is the same as the _________________ of the circular base.

To find the width, imagine taking a can-like cylinder apart with a scissors. When you cut the lateral area, you see that it is equal to the circumference of the can’s top. The circumference of a circle is given by \begin{align*}C = 2 \pi r\end{align*} so the lateral area \begin{align*}L\end{align*} is \begin{align*}L = 2 \pi rh\end{align*}:

\begin{align*}L &= 2 \pi rh\\ &= 2 \pi (5)(24)\\ &= 240 \pi \quad \text{or}\\ & \approx (240)(3.14) = 753.6\end{align*}

Now we can find the area of the entire cylinder using:

\begin{align*}\text{Area = (area of both bases) + (area of lateral side)}\end{align*}

Using exact numbers (with \begin{align*}\pi\end{align*} in them), we get:

\begin{align*}A &= 2(25 \pi) + 240 \pi\\ &= 50 \pi + 240 \pi\\ &= 290 \pi\end{align*}

Using decimal approximations, we get:

\begin{align*}A & = 2(78.5) + 753.6\\ & = 157 + 753.6\\ & = 910.6\end{align*}

The formula we used to find the total surface area can be used for any right cylinder.

Area of a Right Cylinder

The surface area of a right cylinder, with radius \begin{align*}r\end{align*} and height \begin{align*}h\end{align*} is given by

\begin{align*}A = 2B + L\end{align*}

where \begin{align*}B\end{align*} is the area of each base of the cylinder and \begin{align*}L\end{align*} is the lateral area of the cylinder.

Example 1

Use a net to find the surface area of the cylinder:

First draw and label a net for the figure to help you visualize the pieces of the cylinder:

Calculate the area of each base:

\begin{align*}A &= \pi r^2\\ &= \pi(8)^2\\ &= 64 \pi \quad \text{or} \quad \approx (64)(3.14) = 200.96\end{align*}

Calculate the lateral area \begin{align*}L\end{align*}:

\begin{align*}L &= 2 \pi rh\\ &= 2 \pi (8)(9)\\ &= 144 \pi \quad \text{or} \quad \approx (144)(3.14) = 452.16\end{align*}

Find the area of the entire cylinder:

\begin{align*}\text{Area} = \text{(area of both bases)} + \text{(area of lateral side)}\end{align*}

\begin{align*}A &= 2(64 \pi) + 144 \pi &&&& A = 2(200.96) + 452.16\\ &= 128 \pi + 144 \pi && or && \quad = 401.92 + 452.16\\ &= 272 \pi &&&& \quad =854.08\end{align*}

Thus, the total surface area is \begin{align*}272 \pi \ units^2\end{align*} or approximately 854.08 square units.

## Surface Area of a Cylinder Using a Formula

You have seen how to use nets to find the total surface area of a cylinder. The postulate can be broken down to create a general formula for all right cylinders.

\begin{align*}A = 2B + L\end{align*}

Notice that the base \begin{align*}B\end{align*} of any cylinder is:

\begin{align*}B = \pi r^2\end{align*}

The lateral area \begin{align*}L\end{align*} for any cylinder is:

\begin{align*}L &= \text{width of lateral area} \cdot \text{height of cylinder}\\ &= \text{circumference of base} \cdot \text{height of cylinder}\\ &= 2 \pi rh\end{align*}

Putting the two equations together we get:

Area = 2 (area of the _____________________) + ______________________ area

\begin{align*}A &= 2B + L\\ &= 2(\pi r^2) + 2 \pi rh\end{align*}

Factoring out \begin{align*}2 \pi r\end{align*} from the equation gives:

\begin{align*}A = 2 \pi r (r + h)\end{align*}

The Surface Area of a Right Cylinder

A right cylinder with radius \begin{align*}r\end{align*} and height \begin{align*}h\end{align*} can be expressed as:

\begin{align*}A = 2 \pi r^2 + 2 \pi rh \qquad or \qquad A = 2 \pi r (r + h)\end{align*}

You can use the formulas to find the area of any right cylinder.

Example 2

Use the formula to find the surface area of the cylinder:

Write the formula, substitute in the values given above, and then solve:

\begin{align*}A &= 2 \pi r^2 + 2 \pi rh\\ &= 2 \pi( \underline{\; \; \; \; \;\; \; \; \; \;\; \; \; \; \;\; \; \; \; \;}^2) + 2 \pi( \underline{\; \; \; \; \;\; \; \; \; \;\; \; \; \; \;\; \; \; \; \;} )( \underline{\; \; \; \; \;\; \; \; \; \;\; \; \; \; \;\; \; \; \; \;} )\\ &= 450 \pi + 1440 \pi\\ &= 1890 \pi \ in^2 \approx (1890)(3.14) = 5934.6 \ \text{square inches}\end{align*}

The exact area of the cylinder is \begin{align*}1890 \pi \ in^2\end{align*} and the approximate area is \begin{align*}5934.6 \ in^2\end{align*}.

Example 3

Find the surface area of the cylinder:

Write the formula, substitute in the values given above, and then solve: (This time we will try using the other formula for area)

\begin{align*}A &= 2 \pi r (r + h)\\ &= 2 \pi (\underline{\; \; \; \; \;\; \; \; \; \;\; \; \; \; \;\; \; \; \; \;}) \cdot ( \underline{\; \; \; \; \;\; \; \; \; \;\; \; \; \; \;\; \; \; \; \;} + \underline{\; \; \; \; \;\; \; \; \; \;\; \; \; \; \;\; \; \; \; \;} )\\ &=1.5 \pi \cdot 6.75\\ &= 10.125 \pi \ in^2 \approx (10.125)(3.14) = 31.7925 \ in^2\end{align*}

The exact area of the cylinder is \begin{align*}10.125 \pi \ in^2\end{align*} and the approximate area is \begin{align*}31.7925 \ in^2\end{align*}.

Example 4

Find the height of a cylinder that has radius of 4 cm and surface area of \begin{align*}72 \pi \ cm^2\end{align*}.

Write the formula with the given information – you will solve for \begin{align*}h\end{align*}:

\begin{align*}A = 2 \pi r (r + h)\end{align*}

Fill in what you know ( \begin{align*}A\end{align*} and \begin{align*}r\end{align*} ):

\begin{align*}72 \pi = 2 \pi(4) \cdot (4 + h)\end{align*}

Simplify and solve for \begin{align*}h\end{align*}:

\begin{align*}\frac{72 \pi} {8 \pi} &= \frac{8 \pi (4 + h)}{8 \pi}\\ 9 &= 4 + h\\ 5 &= h\end{align*}

The height of the cylinder is 5 cm.

Reading Check:

1. What shapes are the bases of a cylinder?

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

2. When flattened, what shape is the lateral area of a cylinder?

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

3. Describe in words how to find the total surface area of a right cylinder. What information do you need to know? How do you use this information to find the area?

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

## Volume of a Right Cylinder

You have seen how to find the volume of any right prism:

\begin{align*}V = Bh\end{align*}

where \begin{align*}B\end{align*} is the area of the prism’s base and \begin{align*}h\end{align*} is the height of the prism:

As you might guess, right prisms and right cylinders are very similar with respect to volume. A cylinder is just a “prism with round bases.”

• You can think of a cylinder as a prism with bases that are __________________.

One way to develop a formula for the volume of a cylinder is to compare it to a prism. Suppose you divided the prism above into slices that were 1 unit thick:

The volume of each individual slice would be given by the product of the area of the base and the height. Since the height for each slice is 1, the volume of a single slice would be:

\begin{align*}V \text{(single slice)} &= \text{area of base} \cdot \text{height}\\ &= B \cdot 1\\ &= B\end{align*}

Now it follows that the volume of the entire prism is equal to the area of the base multiplied by the number of slices. If there are \begin{align*}h\end{align*} slices, then:

\begin{align*}V \text{(whole prism)} &= B \cdot \text{number of slices}\\ &= Bh\end{align*}

Of course, you already know this formula for volume of a prism. But now you can use the same idea to obtain a formula for the volume of a cylinder.

Look at a cylinder in slices:

Since the height of each slice of the cylinder is 1, each slice has a volume of \begin{align*}B \cdot 1\end{align*}, or \begin{align*}B\end{align*}. Since the base has an area of \begin{align*}\pi r^2\end{align*}, each slice has a volume of \begin{align*}\pi r^2\end{align*} and:

\begin{align*}V (\text{whole cylinder}) &= B \cdot \text{number of slices}\\ &= Bh\\ &= \pi r^2h\end{align*}

This leads to a postulate for the volume of any right cylinder.

Volume of a Right Cylinder

The volume of a right cylinder with radius \begin{align*}r\end{align*} and height \begin{align*}h\end{align*} can be expressed as:

\begin{align*}Volume = \pi r^2h\end{align*}

Example 5

Use the postulate to find the volume of the cylinder:

Write the formula from the postulate. Then substitute in the given values and solve:

\begin{align*}V &= \pi r^2h\\ &= \pi(6.5^2)(14)\\ &= 591.5 \pi \ cm^3 \ \text{or}\\ & \approx (591.5)(3.14) = 1857.31 \ \text{cubic cm}\end{align*}

The exact volume of the given cylinder is \begin{align*}591.5 \pi \ cm^3\end{align*} and the approximate volume is \begin{align*}1857.31 \ cm^3\end{align*}.

Example 6

What is the radius of a cylinder with a height of 10 cm and a volume of \begin{align*}250 \pi \ cm^3\end{align*}?

Write the formula for volume – you will solve for \begin{align*}r\end{align*}:

\begin{align*}V = \pi r^2h\end{align*}

Fill in what you know ( \begin{align*}V\end{align*} and \begin{align*}h\end{align*} ):

\begin{align*}250 \pi = \pi r^2(10)\end{align*}

Simplify and solve for \begin{align*}r\end{align*}:

\begin{align*}\frac{250 \pi}{10 \pi }&= \frac{10 \pi (r^2)}{10 \pi}\\ 25 &= r^2\\ 5 &= r\end{align*}

The radius of the cylinder is 5 cm.

Reading Check:

1. What is the difference between the area of a cylinder and the volume of a cylinder? Describe in your own words.

\begin{align*}\; \; \;\end{align*}

\begin{align*}\; \; \;\end{align*}

\begin{align*}\; \; \;\end{align*}

\begin{align*}\; \; \;\end{align*}

2. What is the difference between finding the exact value of the volume of a cylinder and the approximate volume? How will your answers be different?

\begin{align*}\; \; \;\end{align*}

\begin{align*}\; \; \;\end{align*}

\begin{align*}\; \; \;\end{align*}

\begin{align*}\; \; \;\end{align*}

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Grades:
8 , 9 , 10
Date Created:
Feb 23, 2012
Last Modified:
May 12, 2014
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