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The volume of a solid is the measure of the amount of space inside the solid. It is the three-dimensional equivalent of the area. The volume is measured using cubic units, such as cubic meters, $m_{3}.$ The applet below illustrates the volume of some solids as a blue region by pressing the corresponding buttons.

Different solids may have different formulas to calculate their volumes.

Figure | Name | Formula |
---|---|---|

Cube | $V=s_{3}$ | |

Cone | $V=31 πr_{2}h$ | |

Cylinder | $V=πr_{2}h$ | |

Sphere | $V=34 πr_{3}$ |

Two solids with the same height and the same cross-sectional area at every altitude have the same volume. This means that, as long as their heights are equal, skewed versions of the same solid have the same volume.
### Proof

Informal Justification

If $V_{1}$ and $V_{2}$ are the volumes of the above solids, then they are equal.

$V_{1}=V_{2}$

This principle will be proven by using a set of *identical* coins. Consider a stack in which each of these coins is placed directly on top of each other. Consider also another stack where the coins lie on top of each other, but are not aligned.

The first stack can be considered as a right cylinder. Similarly, the second stack can be considered as an oblique cylinder, which is a *skewed* version of the first cylinder. Because the coins are identical, the cross-sectional areas of the cylinders at the same altitude are the same.

Since the coins are identical, they have the same volume. Furthermore, since the height is the same for both stacks, they both have the same number of coins. Therefore, both stacks — cylinders — have the same volume. This reasoning is strongly based on the assumption that the face of the coins have the same area.

The volume of a solid can be calculated with a formula that depends on the solid's shape. ### Rule

### Prism

The volume of a prism is given by multiplying the area of its base, $B,$ with its height, $h.$
$V=Bh$
### Rule

### Pyramid

The volume of a pyramid is a third of the volume of a prism with the same base. Thus, the formula for the pyramid's volume is:
$V=3Bh .$
### Rule

### Cylinder

A cylinder has a base of a circle. Therefore, the cylinder's volume is calculated by multiplying the area of the circle with the height of the cylinder.
$V=πr_{2}h$
### Proof

Volume of a Cylinder ### Proof

## Volume of a Cylinder

### Rule

### Cone

The volume of a cone is a third of the volume of a cylinder with the same base. Since the cone's base is in the shape of a circle, the formula for the volume of the cone is: $V=3πr_{2}h .$
### Proof

Volume of a Cone ### Proof

## Volume of a Cone

The ratio between the areas is $4π .$ This will be the case for all cross sections of the pyramid and cone. Therefore, Cavalieri's principle implies that the volume of the pyramid is equal to the volume of the cone scaled by the factor $4π .$ The formula for the volume of a pyramid is:
$V_{P}=3Bh .$
As mentioned previously, the base area for the pyramid is expressed as $B=(2r)_{2}.$ $V_{P}=3(2r)_{2}h .$
By multiplying the volume of the pyramid with the scale factor, the formula for the volume of the cone can be found.
The result is the formula for the volume of a cone.
### Rule

### Sphere

A sphere is a perfectly round solid composed of many circles. The formula for the volume of the sphere is given by:
$V=34πr_{3} ,$ where $r$ is the radius of the largest circle in the sphere.

The formula for the volume of a cylinder can be proven by placing a rectangular prism next to it.

Suppose the prism and the cylinder have the same base area and height.

Thus, Cavalieri's principle states that two solids with the same base area and height have the same volume. Since the volume of a prism is given by $V_{P}=Bh,$ the volume of a cylinder can be calculated with the same formula. The base of a cylinder has the shape of a circle. Its area can be expressed as $B=πr_{2}.$ Therefore, the formula for the volume of a cylinder is $V_{C}=πr_{2}h.$

It applies to all cylinders because there is always a prism with the same base area and height.The formula for the volume of a cone can be derived using a pyramid.

Assume, the pyramid and cone have the same height, $h,$ and that the pyramid has a square base, whose side length is twice the radius of the cone's base.

Since the cone's diameter is equal to the side length of the square, the cone will fit inside the pyramid.

The base of the cone is a circle that fits exactly in the base of the pyramid. By studying the figure from below, the ratio between the bases can be determined.

The area of the square is calculated by taking the square of the side length. Further, the area of the circle is given by the square of the radius times $π.$
$A_{S}=(2r)_{2}A_{C}=πr_{2}$
The ratio between the areas can now be calculated by dividing the area of the circle with the area of the square.

$A_{S}A_{C} $

SubstituteExpressions

Substitute expressions

$(2r)_{2}πr_{2} $

CalcPow

Calculate power

$4r_{2}πr_{2} $

SimpQuot

Simplify quotient

$4π $

$V_{C}=4π ⋅3(2r)_{2}h $

CalcPow

Calculate power

$V_{C}=4π ⋅34r_{2}h $

MultFrac

Multiply fractions

$V_{C}=4⋅34⋅πr_{2}h $

SimpQuot

Simplify quotient

$V_{C}=3πr_{2}h $

A solid that is made up of more than one solid is called a composite solid. The individual solids can be combined either by adding or subtracting them from one another. For instance, a hemisphere — half a sphere — can be combined with a cone to make something that resembles a snow cone, or it could be used to hollow out a cylinder.

The volume of a composite solid is either the sum or difference between the volumes of the individual solids, whichever is applicable.

Find the volume of ice cream in the wafer cup.

Assume that the wafer is in the shape of a cone and that the ice cream above the wafer is a hemisphere.

Show Solution

The ice cream can be interpreted as a composite solid, a combination of a cone and a hemisphere.

Thus, the sum of the cone's volume and the hemisphere's volume will be the volume of the ice cream. Let's start with calculating the volume of the cone.

The height of the cone is $4$ inch and the radius is $1$ inch. By substituting the known values, the volume of the cone can be calculated.

$V=3Bh $

Substitute

$B=πr_{2}$

$V=3πr_{2}⋅h $

SubstituteValues

Substitute values

$V=3π⋅1_{2}⋅4 $

$V=34π $

Now, the volume of the hemisphere can be calculated.
Thus, the volume of the hemisphere is $32π in_{3}.$

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