# 10.6: Surface Area of Cylinders

**At Grade**Created by: CK-12

**Practice**Surface Area of Cylinders

Did you have tinker toys when you were little?

Trevor is working on wrapping a container of tinker toys. It is a bit complicated because it is a cylinder. Trevor is on his third attempt, and the woman who bought the tinker toys is getting a bit impatient.

“I’m sorry maam,” Trevor said smiling.

Twice, Trevor did not cut his wrapping paper long enough. The third time, he decides to figure out the surface area of the cylinder first and then cut the wrapping paper.

“I should have done that to begin with,” Trevor thought to himself as he looked at the ruler on the table. Measuring the paper would have been easy had he known the dimensions.

The height of the canister is 18" and the diameter of the cylinder is 6 inches.

Trevor isn’t sure that he has enough information to find the surface area of the cylinder. He stops to think about this for a moment.

**Does Trevor have what he needs? How can he find the surface area of the cylinder?**

**This Concept will teach you how to find the surface area of a cylinder. Make some notes and pay attention, you will see this problem again at the end of the Concept.**

### Guidance

In this Concept, we will learn to find the surface area of cylinders.

**A** *cylinder***is a solid figure that exists in three-dimensional space. A cylinder has two faces that are circles.**

We do not call the side of a cylinder a face because it is curved. We still have to include its area in the total surface area of the cylinder, however.

**The** *surface area***of a cylinder is the total of the area of each circular face and the side of the cylinder.** Imagine a can of soup. The top, bottom, and label around the can would make up the surface area of the can. To find the surface area, we must be able to calculate the area of each face and the side and then add these areas together.

**We will look at two different ways to calculate the surface area of cylinders. One way is to use a net.**

As we’ve said, surface area is the total area for the faces and the side of a cylinder. That means we need to find the area of each face of the cylinder, and then the area of the side. One way to do this is to use a net. **A** *net***is a two-dimensional diagram of a three-dimensional solid.** Imagine you could unroll the soup can so that it is completely flat. You would have something that looks like this.

The shaded circles show the top and bottom faces of the cylinder, and the unshaded rectangle shows the side, as if it were unrolled.

**Can you see how to fold the net back up to make the cylinder?**

With the net, we can see each face of the cylinder more clearly. **To find the surface area, we need to calculate the area for each circle in the net.**

We use the formula \begin{align*}A = \pi r^2\end{align*}

\begin{align*}&\text{bottom face} && \text{top face}\\
A &= \pi r^2 && A = \pi r^2\\
A &= \pi (4)^2 && A = \pi (4)^2\\
A &= 16 \pi && A = 16 \pi\\
A &= 50.24 \ cm^2 && A = 50.24 \ cm^2\end{align*}

**The area of each circular face is 50.24 square centimeters.**

**Now we need to find the area of the side.** The net shows us that, when we “unroll” the cylinder, the side is actually a rectangle. Recall that the formula we use to find the area of a rectangle is \begin{align*}A = lw\end{align*}**the width of the rectangle is the same as the height of the cylinder**. In this case, the height of the cylinder is 8 centimeters.

**What about the length? The length is actually the same as the perimeter of the circle, which we call its circumference.** When we “roll” up the side, it fits exactly once around the circle. To find the area of the cylinder’s side, then, we multiply the circumference of the circle by the height of the cylinder. **We find the circumference of a circle with the formula \begin{align*}C = 2 \pi r\end{align*} C=2πr, and then we multiply it by the height.** Let’s try it.

\begin{align*}C & = 2 \pi r\\
C & = 2 \pi 4\\
C & = 8 \pi\\
C & = 25.12 \times 6 = 150.72 \ cm^2\end{align*}

**Now we know the area of both circular faces and the side. Let’s add them together to find the surface area of the cylinder.**

\begin{align*}& \text{bottom face} \qquad \text{top face} \qquad \quad \text{side} \qquad \qquad \quad \text{surface area}\\
& 50.24 \ cm^2 \quad + \ \ 50.24 \ cm^2 \ + \ 150.72 \ cm^2 \ = \ 251.2 \ cm^2\end{align*}

**The total surface area of the cylinder is 251.72 square centimeters.**

What is the surface area of the figure below?

**The first thing we need to do is draw a net. Get ready to exercise your imagination! It may help to color the top and bottom faces to keep you on track.** Begin by drawing the bottom face. It is a circle with a radius of 7 inches. What shape is the side when we “unroll” the cylinder? It is a rectangle, so we draw a rectangle above the circular base. Lastly, we draw the top face, which is also a circle with a radius of 7 inches.

Here is the net.

Next let’s fill in the measurements for the side and radius of each face so that we can calculate the area of each component. Now we can calculate their areas. Remember to use the correct area and circumference formulas for circles.

\begin{align*}& \text{bottom face} && \text{top face} && \text{side}\\
A &= \pi r^2 && A = \pi r^2 && C = 2 \pi r\\
A &= \pi (7)^2 && A = \pi (7)^2 && C = 2 \pi (7)\\
A &= 49 \pi && A = 49 \pi && C = 14 \pi\\
A &= 153.86 \ in.^2 && A = 153.86 \ in.^2 && C = 43.96 \ in.^2 \times 14 = 615.44 \ in.^2\end{align*}

**Now we add these areas together to find the surface area of the cylinder.**

\begin{align*}153.86 \quad + \quad 153.86 \quad + \quad 615.44 \quad = \quad 923.16 \ in.^2\end{align*}

**Let’s look at what we actually did to find the surface area.**

We used the formula \begin{align*}A = \pi r^2\end{align*}

**We can always draw a net to help us organize information in order to find the surface area of a cylinder. A net helps us see and understand each face of the cylinder.**

**Nets let us see each face so that we can calculate its area. However, we can also use formulas to represent the faces and the side as we find their area.**

You may have noticed in the previous section that the two circular faces always had the same area. This is because they have the same radius. We can therefore calculate the area of the pair of circular faces at once. **We simply double the area formula, which gives us \begin{align*}2 \pi r^2\end{align*} 2πr2.**

We can also combine the measurements for the side into a simpler equation. We need to find the circumference by using the formula \begin{align*}2 \pi r\end{align*}**we can just write \begin{align*}2 \pi rh\end{align*} 2πrh.** When we combine the formula for the faces and for the side we get this formula.

\begin{align*}SA = 2 \pi r^2 + 2 \pi rh\end{align*}

This formula may look long and intimidating, but all we need to do is put in the values for the radius of the circular faces and the height of the cylinder and solve.

*Write this formula down in your notebook.*

Now let’s apply this formula to the example from the previous section.

In this cylinder, \begin{align*}r = 7\end{align*}

\begin{align*}SA & = 2 \pi r^2 + 2 \pi rh\\
SA & = 2 \pi (7)^2 + 2 \pi (7)(14)\\
SA & = 2 \pi (49) + 2 \pi (98)\\
SA & = 98 \pi + 196 \pi\\
SA & = 294 \pi\\
SA & = 923.16 \ in.^2\end{align*}

**As we have already seen, the surface area of this cylinder is 923.16 square inches.**

This formula just saves us a little time. Let’s try another.

What is the surface area of the figure below?

We have all of the measurements we need. Let’s put them into the formula and solve for surface area, \begin{align*}SA\end{align*}

\begin{align*}SA & = 2 \pi r^2 + 2 \pi rh\\
SA & = 2 \pi (3.5)^2 + 2 \pi (3.5) (28)\\
SA & = 2 \pi (12.25) + 2 \pi (98)\\
SA & = 24.5 \pi + 196 \pi\\
SA & = 220.5 \pi\\
SA & = 692.37 \ cm^2\end{align*}

**This cylinder has a surface area of 692.37 square centimeters.**

Try a few of these on your own. Find the surface area of each cylinder.

#### Example A

A cylinder with a radius of 5 ft and a height of 10 ft

**Solution: \begin{align*}471\end{align*} 471 sq. ft.**

#### Example B

A cylinder with a radius of 7 in and a height of 12 in

**Solution: \begin{align*}835.24\end{align*} 835.24 sq. in**

#### Example C

A cylinder with a diameter of 4 m and a height of 5 m

**Solution: \begin{align*}87.92\end{align*} 87.92 sq. m**

Here is the original problem again. Then find the surface area of the cylinder.

Trevor is working on wrapping a container of tinker toys. It is a bit complicated because it is a cylinder. Trevor is on his third attempt, and the woman who bought the tinker toys is getting a bit impatient.

“I’m sorry maam,” Trevor said smiling.

Twice, Trevor did not cut his wrapping paper long enough. The third time, he decides to figure out the surface area of the cylinder first and then cut the wrapping paper.

“I should have done that to begin with,” Trevor thought to himself as he looked at the ruler on the table. Measuring the paper would have been easy had he known the dimensions.

The height of the canister is 18" and the diameter of the cylinder is 6 inches.

Trevor isn’t sure that he has enough information to find the surface area of the cylinder. He stops to think about this for a moment.

**We can use the formula for finding the surface area of a cylinder to help us to find the surface area of this cylinder.**

\begin{align*}SA=2\pi rh+2 \pi r^2\end{align*}

**Now we can take the dimensions and then substitute them into the formula. The first thing to notice is that the diameter of the canister has been given and we need the radius of the cylinder. The radius is one-half of the diameter.**

\begin{align*}6 \div 2 = 3 \ inches\end{align*}

**Now we can substitute them into the formula.**

\begin{align*}SA & = 2(3.14)(3)(18)+2(3.14)(3^2)\\
SA & = 339.12+56.52 \\
SA & = 395.64 \ sq.inches\end{align*}

**We can divide this measurement by 12 and we will know how many square feet of wrapping paper will be needed.**

\begin{align*}395.64 \div 12 = 32.97 \ \text{or} \ 33 \ sq. feet\end{align*}

### Vocabulary

Here are the vocabulary words in this Concept.

- Cylinder
- a three-dimensional figure with two circular bases.

- Surface Area
- the measurement of the outside of a three-dimensional figure.

- Net
- a two-dimensional representation of a three-dimensional figure.

### Guided Practice

Here is one for you to try on your own.

Mrs. Johnson is wrapping a cylindrical package in brown paper so that she can mail it to her son. The package is 22 centimeters tall and 11 centimeters across. How much paper will she need to cover the package?

**Answer**

The picture clearly shows us the height and diameter of the cylinder, so let’s use the formula for finding the surface area. But be careful—we have been given the diameter, not the radius. We need to divide it by 2 to find the radius: \begin{align*}11 \div 2 = 5.5\end{align*}

\begin{align*}SA & = 2 \pi r^2 + 2 \pi rh\\
SA & = 2 \pi (5.5^2) + 2 \pi (5.5) (22)\\
SA & = 2 \pi (30.25) + 2 \pi (121)\\
SA & = 60.5 \pi + 242 \pi\\
SA & = 302.5 \pi\\
SA & = 949.85 \ cm^2\end{align*}

**Mrs. Johnson will need 949.85 square centimeters of brown paper in order to wrap the entire package.**

### Video Review

Here is a video for review.

- This is a video on surface area of cylinders.

### Practice

Directions: Find the surface area of each cylinder given its height and radius.

1. \begin{align*}r = 1 \ m, \ h = 3 \ m\end{align*}

2. \begin{align*}r = 2 \ cm, \ h = 4 \ cm\end{align*}

3. \begin{align*}r = 6 \ in, \ h = 10 \ in\end{align*}

4. \begin{align*}r = 4 \ in, \ h = 6 \ in\end{align*}

5. \begin{align*}r = 5 \ in, \ h = 10 \ in\end{align*}

6. \begin{align*}r = 8 \ ft, \ h = 6 \ ft\end{align*}

7. \begin{align*}r = 10 \ m, \ h = 15 \ m\end{align*}

8. \begin{align*}r = 9 \ cm, \ h = 12 \ cm\end{align*}

9. \begin{align*}r = 6 \ m, \ h = 8 \ m\end{align*}

10. \begin{align*}r = 2 \ cm, \ h = cm\end{align*}

Directions: Find the surface area of each cylinder given its height and diameter.

11. \begin{align*}d = 8 \ m, \ h = 11 \ m\end{align*}

12. \begin{align*}d = 10 \ in, \ h = 14 \ in\end{align*}

13. \begin{align*}d = 8 \ cm, \ h = 10 \ cm\end{align*}

14. \begin{align*}d = 12 \ m, \ h = 15 \ m\end{align*}

15. \begin{align*}d = 15 \ in, \ h = 20 \ in\end{align*}

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