Topics
7. Surface Area & Volume
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Grade 7

7.02 Surface area of right cylinders

Lesson
Practice

Surface area of right cylinders

A cylinder and its net. The net is made up of a rectangle, and 2 identical circles. The circles are tangent to the lengths of the rectangle.

A cylinder has three faces: two identical circular bases and a curved surface that joins the two bases together. The surface area of a cylinder is the sum of the areas of these three faces.

Let's see how a cylinder would look if we open it up and view its net.

Exploration

Drag the slider to unfold the cylinder. Then drag the next slider to rotate the circle.

Loading interactive...

  1. What type of shape did the outside of the cylinder become when it was unfolded?

  2. How many circles are in the net?

  3. Why does the last stage show us that the base circle can roll across the rectangle?

  4. What is the relationship between the circles and the rectangle?

  5. Write the formula for the surface area of a cylinder.

From the applet, we can see that there are three parts to a cylinder's surface area - two circles and a rectangle. One side length of the rectangle is equal to the height of the cylinder and the other side length is the circumference of the base circle. The area of the curved part of a cylinder is 2\pi rh, where r is the radius and h is the height.

The image shows a cylinder and its net with with dimensions. Ask your teacher for more information.

To find the surface area of the whole cylinder, we need to add the area of the top and bottom circles to the area of the curved part.

\displaystyle \text{Surface area of a cylinder}\displaystyle =\displaystyle 2\cdot \left(\text{Area of circular base}\right)+\text{area of curved face}
\displaystyle =\displaystyle 2\cdot \left(\pi \cdot r^2\right)+2 \pi r \cdot h

Now that we have the area of all of the parts, we can put them together to get the total surface area.

\displaystyle SA_\text{cylinder}=2\pi r^{2}+2\pi rh
\bm{r}
the radius of the cylinder
\bm{h}
the height of the cylinder

Examples

Example 1

Consider the following cylinder and it's corresponding net:

A cylinder and its net. The cylinder has a radius of 3 meters and height of 5 meters. The net is composed of 2 identical circles with a radius of 3 meters, and a rectangle with a width of 5 meters.
a

Find the area of one of the circular faces of the cylinder. Use 3.14 for \pi and round your answer to two decimal places.

Worked Solution
Create a strategy

Remember that the area of a circle is A = \pi r^{2}

Apply the idea
\displaystyle A\displaystyle =\displaystyle 3.14 \cdot 3^{2}Substitute the values
\displaystyle =\displaystyle 28.26Evaluate

The area of a circle of the cylinder is 28.26\operatorname{ m}^{2}.

b

Find the area of the curved face. Use 3.14 for \pi and round your answer to two decimal places.

Worked Solution
Create a strategy

The area of the curved face of the cylinder will be the area of the middle rectangle, which makes up the side of the cylinder.

The area of the rectangle is equal to the product of its length and width.

The width of the rectangle is equal to the given height of the cylinder. What will the length of the rectangle be?

Apply the idea

The length of the rectangle is equal to the circumference of the circular base which is 2 \pi r. Therefore, the surface area of the curved face of a cylinder is 2 \pi r \cdot h.

The cylinder has a height of 5 \operatorname { m}, and a radius of 3 \operatorname { m}.

\displaystyle \text{Curved surface area}\displaystyle =\displaystyle 2 \cdot 3.14 \cdot 3 \cdot 5Substitute the values
\displaystyle =\displaystyle 94.20Evaluate

The curved surface area is 94.20 \operatorname{ m}^{2}.

Reflect and check

We may get slightly different values for surface area depending on the approximation we use for \pi. Common approximations for \pi include 3.14 and \dfrac{22}{7}.

If we had used \dfrac{22}{7}, our calculation would have been:

\displaystyle \text{Curved surface area}\displaystyle =\displaystyle 2 \cdot \dfrac{22}{7} \cdot 3 \cdot 5Substitute the values
\displaystyle =\displaystyle 94.29Evaluate

This is slightly larger than the number we got using 3.14 to approximate \pi. Depending on the context this may or may not matter. For the most accurate calculations you can use the \pi button on your calculator. This will keep more digits of \pi than any of the other approximations.

c

Find the total surface area of the cylinder. Round your answer to two decimal places.

Worked Solution
Create a strategy

The surface of a cylinder is made up of two circular faces on the top and bottom and a rectangular face that wraps around the curved surface of the cylinder.

Apply the idea

The surface area of the cylinder can be calculated by adding the areas of each part:

\displaystyle \text{Surface area of a cylinder}\displaystyle =\displaystyle \text{Area of } 2 \text{ circular bases} + \text{Area of rectangular piece}
\displaystyle SA\displaystyle =\displaystyle 2\cdot 28.26\operatorname{ m}^{2} + 94.20 \operatorname{ m}^{2}Substitute the values from parts (a) and (b)
\displaystyle =\displaystyle 150.72 \operatorname{ m}^{2}Evaluate

The surface area of the cylinder is 150.72 \operatorname{ m}^{2}.

Idea summary

A cylinder is a 3D shape much like a prism with two identical circular bases and a curved surface that joins the two bases together.

The surface area of the cylinder can be calculated by totaling the area of the parts:

\text{Surface area of a cylinder} = \text{Area of } 2 \text{ circular ends} + \text{Area of rectangular piece}

The formula for surface area of a cylinder is:

\displaystyle SA_\text{cylinder}=2\pi r^{2}+2\pi rh
\bm{r}
the radius of the cylinder
\bm{h}
the height of the cylinder

Outcomes

7.MG.1

The student will investigate and determine the volume formulas for right cylinders and the surface area formulas for rectangular prisms and right cylinders and apply the formulas in context.

7.MG.1b

Develop the formulas for determining the surface area of rectangular prisms and right cylinders and solve problems, including those in contextual situations, using concrete objects, two- dimensional diagrams, nets, and formulas.

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