Cylinders
Although a cylinder is not a prism - because it has a curved surface between the ends - just like a prism we can use its net to help us develop a rule for finding its surface area.
Let's see how it would look if we open up a cylinder to view its net.
When the net is unfolded, the curved surface is shown to be a rectangle. We can see that there are three parts to a cylinder's surface area - two circles and a rectangle.
| Surface area of a cylinder | $=$= | area of $2$2 circular ends + area of $1$1 rectangular piece |
| $=$= | $\left(2\times\pi r^2\right)+\left(L\times W\right)$(2×πr2)+(L×W) |
We know that the length of the rectangle is the height, $h$h, of the cylinder. By rotating the circle on top of the rectangle, we can see how the circumference of the circle is equal to the width of the rectangular piece? The circumference of a the circle is given by $2\pi r$2πr, so we have:
$A=2\pi r^2+2\pi rh$A=2πr2+2πrh
Questions can involve fractions of cylinders: for example, a water trough might be made from exactly half of a cylinder. We might be told that a cylinder is open, which would mean it has no circles as part of its surface area. Or, it might have one end closed, such as a water tank collecting rain. So it is important that we understand exactly how the formula above works so that we can use it in these more complicated scenarios.
Worked Example
Example 1
Calculate the surface area of the closed half cylinder below, giving your answer to $1$1 decimal place.
Think: The surface area is composed of half of a closed cylinder and an extra rectangle on top.
Do: The surface area of a cylinder = $2\pi r^2+2\pi rh$2πr2+2πrh, so
The surface area of half a cylinder = $\pi r^2+\pi rh$πr2+πrh $=\pi\times3.6^2+\pi\times10\times3.6=153.8123$=π×3.62+π×10×3.6=153.8123...
The rectangle has its length equal to the height of the cylinder, and its width is equal to the diameter of the cylinder's ends.
We have the area of the rectangle $=10\times7.2=72$=10×7.2=72
So the total surface area is $153.8123$153.8123... $+72=225.8$+72=225.8 square units ($1$1 d.p.)
Practice Questions
QUESTION 1
Consider the following cylinder with a height of $35$35 cm and base radius of $10$10 cm. Find the surface area of the cylinder.
A vertical cylinder with a dashed circle at the bottom indicating the base, which is hidden from view. A solid line circles the top, representing the visible base. Two measurements are shown: a horizontal line from the center of the circle to the circumference measuring $10$10cm above the top base, implying the radius, and a vertical dimension line adjacent to the right side of the cylinder measuring $35$35cm indicating the height.
Round your answer to two decimal places.
QUESTION 2
A cylindrical can of radius $7$7 cm and height $10$10 cm is open at one end. What is the external surface area of the can correct to two decimal places?
Question 3
Consider the cylinder shown in the diagram below.
Find the surface area of the cylinder in square centimetres.
Round your answer to one decimal place.
Use your answer from part (a) to find the surface area of the cylinder in square millimetres?