 # Volume and Surface Area of Three-dimensional Figures

## Three-dimensional Figures

The most common three-dimensional figures are: prisms, cylinders, pyramids, cones and spheres. They all have two important qualities: volume and surface area. Here is a list containing the figures and their properties.

### Prism Example 1

A prism has a triangle with the area $B=\text{}12\text{}\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{2}$ as its base, and height $h=\text{}5\text{}\phantom{\rule{0.17em}{0ex}}\text{cm}$. Calculate the volume of the prism.

You’ll find the volume by multiplying the area of the base $B$ with the height $h$:

 $\text{Volume}=B\cdot h=12\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{2}\cdot 5\phantom{\rule{0.17em}{0ex}}\text{cm}=60\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{3}.$

### Cylinder Example 2

A cylinder has the radius $r=\text{}2.0\text{}\phantom{\rule{0.17em}{0ex}}\text{cm}$ and height $h=\text{}8.0\text{}\phantom{\rule{0.17em}{0ex}}\text{cm}$. Calculate the volume and the surface area of the cylinder.

### Volume

The formula for the volume of a cylinder is $V=B\cdot h$. You have the height $h$, but have to find the base area $B$. Since the base area in a cylinder is a circle, you’ll have to find the area of a circle:

 $B=\pi {r}^{2}=\pi \cdot 2.0\phantom{\rule{0.17em}{0ex}}\text{cm}\cdot 2.0\phantom{\rule{0.17em}{0ex}}\text{cm}\approx 12.6\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{2}.$

Now you’ll find the volume of a cylinder by multiplying the base area $B$ with the height $h$:

$\begin{array}{llll}\hfill \text{Volume}& =B\cdot h\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 12.6\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{2}\cdot 8.0\phantom{\rule{0.17em}{0ex}}\text{cm}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 101\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{3}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{lll}\hfill \text{Volume}=B\cdot h\approx 12.6\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{2}\cdot 8.0\phantom{\rule{0.17em}{0ex}}\text{cm}\approx 101\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{3}.& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$

### Surface Area

To find the surface area of a cylinder, you add together the areas of all the sides. The green rectangle you see on the drawing comes from cutting and folding out the walls of the cylinder. First, you calculate the area of the top and bottom of the cylinder (they have the same area), and then the area of the rectangle which is the height. The length of the rectangle is the circumference of the circle (see the figure):

The surface area of a cylinder then becomes

### Pyramid Example 3

A pyramid has a triangle with area $B=\text{}9\text{}\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{2}$ as its base. The height $h$ of the pyramid is $\text{}4\text{}\phantom{\rule{0.17em}{0ex}}\text{cm}$. Calculate the volume of the pyramid.

You’ll find the volume of the pyramid by multiplying the area of the base area $B$ with the height $h$, and the dividing by 3:

 $\text{Volume}=\frac{B\cdot h}{3}=\frac{9\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{2}\cdot 4\phantom{\rule{0.17em}{0ex}}\text{cm}}{3}=12\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{3}.$

### Cone Example 4

A cone has a circle with radius $r=\text{}3.0\text{}\phantom{\rule{0.17em}{0ex}}\text{cm}$ as its base. The height $h$ of the cone is $\text{}7.0\text{}\phantom{\rule{0.17em}{0ex}}\text{cm}$ and side $s=\text{}9.0\text{}\phantom{\rule{0.17em}{0ex}}\text{cm}$. Calculate the volume and the surface area of the cone.

### Volume

The volume has the formula $V=B\cdot h$, so you first need to find the area of the base $B$. The area of the circle is:

 $B=\pi {r}^{2}=\pi \cdot 3.0\phantom{\rule{0.17em}{0ex}}\text{cm}\cdot 3.0\phantom{\rule{0.17em}{0ex}}\text{cm}\approx 28.3\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{2}.$

Then you’ll find the volume of the cone by multiplying the area of the base $B$ with the height $h$, and then divide by 3: $\begin{array}{llll}\hfill \text{Volume}& =\frac{B\cdot h}{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{28.3\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{2}\cdot 7.0\phantom{\rule{0.17em}{0ex}}\text{cm}}{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 66.0\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{3}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

### Surface Area

You’ll find the surface area by using the formula for the surface area of a cone. Here you can put your values directly in to the formula and calculate.

### Sphere Example 5

A sphere has radius $\text{}5.0\text{}\phantom{\rule{0.17em}{0ex}}\text{cm}$. Calculate the volume and the surface area of the sphere.

### Volume

You1’ll find the volume by using the formula for the volume of a sphere: $\begin{array}{llll}\hfill \text{Volume}& =\frac{4\pi {r}^{3}}{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{4\cdot \pi \cdot 5.0\phantom{\rule{0.17em}{0ex}}\text{cm}\cdot 5.0\phantom{\rule{0.17em}{0ex}}\text{cm}\cdot 5.0\phantom{\rule{0.17em}{0ex}}\text{cm}}{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 524\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{3}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

### Surface Area

You’ll find the surface area by using the formula for the surface area of a sphere: