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10.4: Surface Area of Cylinders

Created by: CK-12

Introduction

The Tinker Toys

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 ma'am,” 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 thinks to himself as he looks 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 lesson 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 lesson.

What You Will Learn

By the end of this lesson you will be able to use the following skills:

  • Recognize the surface area of cylinders as the sum of areas of faces using nets.
  • Find surface areas of cylinders using formulas.
  • Find heights of cylinders given surface areas and another dimension.
  • Solve real-world problems involving surface areas of cylinders.

Teaching Time

I. Recognize the Surface Area of Cylinders as the Sum of Areas of Faces using Nets

In this lesson 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. However, we still have to include its area in the total surface area of the cylinder.

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 of 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 A = \pi r^2 to find the area of a circle. If we know the radius or diameter of each circle, we can calculate its area. Look closely again at the cylinder above. The two circular faces are congruent, so they must have the same radius and diameter. Let’s calculate the area of each face.

&\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

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 A = lw. For cylinders, 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 C = 2 \pi r, and then we multiply it by the height. Let’s try it.

C & = 2 \pi r\\C & = 2 \pi 4\\C & = 8 \pi= 25.12\\C & = 25.12 \times 8 = 200.96 \ cm^2

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.

& \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 \ + \ 200.96 \ cm^2 \ = \ 301.44 \ cm^2

The total surface area of the cylinder is 301.44 square centimeters.

Example

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.

& \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

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

153.86 \quad + \quad 153.86 \quad + \quad  615.44 \quad = \quad  923.16 \ in.^2

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

We used the formula A = \pi r^2 to find the area of the top and bottom faces. We used the formula C = 2 \pi r to find the circumference of the circular base and multiplied this by the height to find the area of the side. When we add these together, we get a surface area of 923.16 square inches for this cylinder. Nice work!

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. We can also use a formula to find the area of cylinders.

II. Find Surface Areas of Cylinders using Formulas

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 2 \pi r^2.

We can also combine the measurements for the side into a simpler equation. We need to find the circumference by using the formula 2 \pi r, and then we multiply this by the height of the cylinder. Therefore we can just write 2 \pi rh. When we combine the formula for the faces and for the side we get this formula.

SA = 2 \pi r^2 + 2 \pi rh

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, r = 7 inches and h = 14 inches. We simply put these numbers into the formula and solve for surface area. Let’s try it.

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

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 problem:

Example

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, SA.

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

This cylinder has a surface area of 692.37 square centimeters.

Try a few of these on your own.

10F. Lesson Exercises

Find the surface area of each cylinder.

  1. A cylinder with a radius of 5 ft and a height of 10 ft
  2. A cylinder with a radius of 7 in and a height of 12 in
  3. A cylinder with a diameter of 4 m and a height of 5 m

Take a few minutes to check your work with a peer.

III. Find Heights of Cylinders Given Surface Areas and Another Dimension

Sometimes we may be given the surface area of a cylinder and need to find its height. We simply fill the information we know into the surface area formula. This time, instead of solving for SA, we solve for h, the height of the cylinder. Let’s look at an example to see how this works.

Example

The surface area of a cylinder with a radius of 3 inches is 78 \pi square inches. What is the height of the cylinder?

We know that the radius of the cylinder is 3 inches. We also know that the surface area of the cylinder is 78 \pi square inches. Sometimes a problem will include pi in the amount because it is more precise. We’ll see that this also makes our calculations easier. All we have to do is put 78 \pi into the formula in place of SA. Then we can use the formula to solve for the height, h.

SA & = 2 \pi r^2 + 2 \pi rh\\78 \pi & = 2 \pi (3^2) + 2 \pi (3) h\\78 \pi & = 2 \pi (9) + 6 \pi h\\78 \pi & = 18 \pi + 6 \pi h\\78 \pi - 18 \pi & = 6 \pi h \qquad \qquad \qquad \text{Subtract} \ 18 \pi \ \text{from both sides.}\\60 \pi & = 6 \pi h\\60 \pi \div 6 \pi & = h \qquad \qquad \quad \qquad \text{Divide both sides by} \ 6 \pi.\\10 \ in. & = h

We used the formula to find that a cylinder with a radius of 3 inches and a surface area of 78 \pi has a height of 10 inches.

We can check our work by putting the height into the formula and solving for surface area.

SA & = 2 \pi (3^2) + 2 \pi (3) (10)\\SA & = 2 \pi (9) + 2 \pi (30)\\SA & = 18 \pi + 60 \pi\\SA & = 78 \pi

This is the surface area given in the problem, so our answer is correct. Let’s look at another example.

Example

Mrs. Javitz bought a container of oatmeal in the shape of the cylinder. If the container has a radius of 5 cm and a surface area of 170 \pi, what is its height?

Once again, we have been given the surface area as a function of pi. That's no problem, we simply put the whole number, 170 \pi, in for SA in the formula. We also know that the container has a radius of 5 centimeters, so we can use the formula to solve for h, the height of the container.

SA & = 2 \pi r^2 + 2 \pi rh\\170 \pi & = 2 \pi (5^2) + 2 \pi (5) h\\170 \pi & = 2 \pi (25) + 10 \pi h\\170 \pi & = 50 \pi + 10 \pi h\\170 \pi - 50 \pi & = 10 \pi h \quad \qquad \qquad \text{Subtract} \ 50 \pi \ \text{from both sides.}\\120 \pi & = 10 \pi h\\120 \pi \div 10 \pi & = h \qquad \qquad \qquad \ \ \text{Divide both sides by} \ 10 \pi.\\12 \ cm & = h

The oatmeal container must have a height of 12 centimeters.

Now try a few of these on your own.

10F. Lesson Exercises

Find the height in each problem.

  1. SA = 48 \pi, \ r = 3 \ in
  2. SA = 40 \pi, \ r = 2 \ in
  3. SA = 80 \pi, \ r = 4 \ m

Take a few minutes to check your work with a partner. Correct any errors and then move on to the next section.

IV. Solve Real-World Problems Involving Surface Area of Cylinders

We have learned two ways to find surface area: drawing a net or using a formula. We can use either of these methods to solve word problems involving surface area. Nets may be especially useful if the problem does not provide an image of the cylinder. Let’s practice using what we have learned.

Example

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?

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: 11 \div 2 = 5.5. Now we have the radius and height, so we can put these in for the appropriate variables in the formula.

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

Mrs. Johnson will need 949.85 square centimeters of brown paper in order to wrap the entire package. Let’s try another.

Example

Travis is making lanterns by covering glasses in colored paper. Each glass has a radius of 2 inches and a height of 6.5 inches. If he covers the sides and bottom of 6 glasses, how many square inches of paper will he need?

Read the problem carefully! What is it asking us to find? We need to find the surface area of only the bottom face and side of the cylinder, not the top, because Travis is not wrapping the entire cylinder in paper. Notice also that we need to find how much paper he will need for six glasses. This problem is a bit more complicated, so it may help to draw a net. In your drawing, shade the parts covered in paper: bottom of the glass and the side. (The diagram below has all of the sides shaded. Which part of the net should you remove the shading from to make sure it is correct?)

With a net, we can see clearly which face represents the bottom and which represents the unpapered top. Now let’s fill in all of the measurements and solve for the area of the bottom and side of the glass.

& \text{bottom} && \text{side}\\A & = \pi r^2 && A = 2 \pi rh\\A &= \pi (2)^2 && A = 2 \pi (2) (6.5)\\A &= 4 \pi && A = 2 \pi (13)\\A &= 12.56 \ in.^2 && A = 26 \pi\\&&&A = 81.64 \ in.^2 &&

Now we add the area of each component to find the total surface area that Travis will cover with paper (the side and bottom of the glass).

12.56 \ in.^2 \quad + \quad 81.64 \ in.^2 \quad = \quad 94.2 \ in.^2

Travis will need 94.2 square inches of canvas to cover one glass. But we’re not done yet! Remember that the problem asked us to find how much paper he will need to cover six glasses. We need to multiply the surface area of one glass by 6.

94.2 \ in.^2 \times 6 \ = \ 565.2 \ in.^2

Travis will need 565.2 square inches of paper to cover the sides and bottom of all six glasses.

Remember to read word problems carefully and be sure you understand what they are asking them to find. Now let’s go back to the problem in the introduction and solve it.

Real–Life Example Completed

The Tinker Toys

Here is the original problem again. Reread it and underline any important information.

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 ma'am,” 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 thinks to himself as he looks 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.

SA=2\pi rh+2 \pi r^2

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.

6 \div 2 = 3 \ inches

Now we can substitute them into the formula.

SA & = 2(3.14)(3)(18)+2(3.14)(3^2)\\SA & = 339.12+56.52 \\SA & = 395.64 \ sq.inches

We can divide this measurement by 144 and we will know how many square feet of wrapping paper will be needed.

395.64 \div 144 = 2.7475 \ \text{or} \ 2.75 \ sq. feet

Vocabulary

Here are the vocabulary words found in this lesson.

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.

Technology Integration

Khan Academy Cylinder Volume and Surface Area

  1. http://www.mathplayground.com/mv_surface_area_cylinders.html – This is a video on how to find the surface area of a cylinder.

Time to Practice

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

1. r = 1 \ m, \ h = 3 \ m

2. r = 2 \ cm, \ h = 4 \ cm

3. r = 6 \ in, \ h = 10 \ in

4. r = 4 \ in, \ h = 6 \ in

5. r = 5 \ in, \ h = 10 \ in

6. r = 8 \ ft, \ h = 6 \ ft

7. r = 10 \ m, \ h = 15 \ m

8. r = 9 \ cm, \ h = 12 \ cm

9. r = 6 \ m, \ h = 8 \ m

10. r = 2 \ cm, \ h = cm

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

11. d = 8 \ m, \ h = 11 \ m

12. d = 10 \ in, \ h = 14 \ in

13. d = 8 \ cm, \ h = 10 \ cm

14. d = 12 \ m, \ h = 15 \ m

Directions: Figure out the radius of each cylinder given the height and the surface area.

15. h = 5 \ in, \ SA = 87.92 \ sq. in.

16. h = 5 \ in, \ SA = 150.72 \ sq. in

17. h = 7 \ m, \ SA = 489.84 \ sq. m

18. h = 7 \ cm, \ SA = 276.32 \ sq. cm.

19. h = 10 \ in, \ SA = 471 \ sq. in.

20. h = 12 \ m, \ SA = 1186.92 \ sq. m.

21. h = 8 \ m, \ SA = 301.44 \ sq. m.

22. h = 9 \ m, \ SA = 703.36 \ sq. m.

23. h = 11 \ in, \ SA = 502.4 \ sq. in.

24. h = 10 \ in, \ SA = 602.88 \ sq. in

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Date Created:

Feb 22, 2012

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Mar 07, 2014
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CK.MAT.ENG.SE.1.Math-Grade-7.10.4

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