Question

A cylinder has a height of $10$ and a radius of $3.$ Determine the total surface area, including the two ends, of the cylinder.

Answer 1

Answer:

$78\, \pi$.

Step-by-step explanation:

A cylinder of height $h$ and radius $r$ could be unrolled into:

• two identical circles, one for the top and one for the bottom, and
• one rectangle, for the side.

The radius of each circle is the same as the radius of the cylinder, $r$.

The height of the rectangle is the same as the height of the cylinder, $h$.

The width of this rectangle would be $(2\, \pi\, r)$, same as the circumferences of the two circles. The reason is that when this rectangle is rolled to build the cylinder, the top and bottom edges of the rectangle would wrap around the circumference of the two circles.

The area of each circle would be $\pi\, r^{2}$. Note that there are two such circles, one at the top of the cylinder and one at the bottom.

With a height of $h$ and a width of $(2\, \pi\, r)$, the area of the rectangle on the side would be $(2\, \pi\, r)\, h$. Add these areas up to find the total surface area of this cylinder:

$\begin{aligned}& \text{(Surface Area of the Cylinder)} \\ =\; & (\text{Surface area of circle on the top}) \\ & + (\text{Surface area of circle at the bottom}) \\ & + (\text{Surface area of rectangle on the side}) \\ =\; & \pi\, r^{2} + \pi\, r^{2} + (2\, \pi\, r)\, h \\ =\; & 2\, \pi\, r^{2} + (2\, \pi\, r)\, h \end{aligned}$.

For the cylinder in this question, $h = 10$ whereas $r = 3$. The surface area of this cylinder would be:

$\begin{aligned}& 2\, \pi\, r^{2} + (2\, \pi\, r) \, h \\ =\; & 2 \,\pi \times 3^{2} + (2\, \pi\times 3) \times 10 \\ =\; & 78\, \pi\end{aligned}$.

Answer 2

Check the picture below.

$\textit{surface area of a cylinder}\\\\ SA=2\pi r(h+r)~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=3\\ h=10 \end{cases}\implies \begin{array}{llll} SA=2\pi (3)(10+3) \\\\\\ SA=6\pi (13)\implies SA=78\pi \\\\\\ SA\approx 245.04 \end{array}$