8.04 Volume and capacity
Volume is the space taken up by a 3D solid and is measured in units cubed. Capacity is the amount the solid could hold.
cubic millimetres = mm3
(picture a cube with side lengths of $1$1 mm each - this one is pretty small!)
cubic centimetres = cm3
(picture a cube with side lengths of $1$1 cm each - about the size of a dice)
cubic metres = m3
(picture a cube with side lengths of $1$1 m each - what could be this big?)
To convert to capacity
$1$1cm3 = $1$1 mL
Prisms
A prism has a uniform cross-section (determined by its bases). Prisms occur very commonly in packaging of grocery items, and finding the volume of these contributes to the design, shape and size of packaging and product - all of these affect the price that we pay!
To experiment with how the volume of a triangular prism is affected by its base (cross section) and height, or to see how the volume of a rectangular prism is affected by its base, width and height, try the following mathlet. You can vary the dimensions by moving the sliders and expose the volume by checking the checkbox.
$\text{Volume }=\text{Area of cross section }\times\text{height }$Volume =Area of cross section ×height
$V=Ah$V=Ah
In the case of a triangular prism, the volume is calculated by:
$\text{Area of the triangle }\times\text{height of the prism }$Area of the triangle ×height of the prism .
For a trapezium based prism, the volume is calculated by:
$\text{Area of the trapezium }\times\text{height of the prism}$Area of the trapezium ×height of the prism .
For the volume of a prism like the one to the right the volume is calculated by:
$\text{Area of the L shape }\times\text{height of the prism }$Area of the L shape ×height of the prism .
Rectangular prisms
The cross section or base of a rectangular prism is a rectangle so the area of the base is always
$\text{Length }\times\text{Width }$Length ×Width
$\text{Volume of rectangular prism }=\text{Length }\times\text{Width }\times\text{Height }$Volume of rectangular prism =Length ×Width ×Height
$V=lwh$V=lwh
Cylinders
A cylinder is very similar to a prism (except for the rounded cross section or base) and the volume can be found using a similar formula where the base is a circle.
| $\text{Volume of cylinder}$Volume of cylinder | $=$= | $\text{Area of base }\times\text{height of prism }$Area of base ×height of prism |
| $=$= | $\pi r^2\times h$πr2×h | |
| $=$= | $\pi r^2h$πr2h |
Exploration
You are at the local hardware store to buy a can of paint. After settling on one product, the salesman offers to sell you a can that is either double the height or double the radius (your choice) of the one you had decided on for double the price. Assuming all cans of paint are filled to the brim, is it worth taking up his offer?
If so, would you get more paint for each dollar if you chose the can that was double the radius or the can that was double the height?
Since the volume of a cylinder is given by the formula $\pi r^2h$πr2h, if the height doubles, the volume becomes $\pi r^2\times2h=2\pi r^2h$πr2×2h=2πr2h (ie the volume increases two-fold).
Whereas if the radius doubles, the volume becomes $\pi\left(2r\right)^2h=4\pi r^2h$π(2r)2h=4πr2h (ie the volume increases four-fold).
That is, the volume increases by double the amount when the radius is doubled compared to when the height is doubled.
To see how changes in height and radius affect the volume of a can to different extents, try the following interactive. You can vary the height and radius by moving the sliders around.
Practice questions
Question 1
Find the volume of the cube shown.
QUESTION 2
Find the volume of the triangular prism shown.
QUESTION 3
Find the volume of the prism by finding the base area first.
Pyramids
A pyramid is formed when the vertices of a polygon are projected up to a common point (called an apex). A right pyramid is formed when the apex is directly above the centre of the base.
To prove and derive the formula for the volume of a prism is beyond the skills we currently have at this level, but we can describe a more simple way to consider it.
Exploration
Think about a cube, with side length $s$sunits. Now let's divide the cube up into $6$6 simple pyramids by joining all the vertices to the midpoint of the cube.
This creates $6$6 square based pyramids with the base equal to the other faces of the cube, and height equal to half the length of the side.
$\text{Volume of the cube }=s^3$Volume of the cube =s3
$\text{Volume of one of the pyramids }=\frac{s^3}{6}$Volume of one of the pyramids =s36
Now, let's think about a rectangular prism, that is, half the cube. This rectangular prism has the same base as the pyramid and the same height as the pyramid.
Now the volume of this rectangular prism is $l\times b\times h=\frac{s\times s\times s}{2}$l×b×h=s×s×s2= $\frac{s^3}{2}$s32
We know that the volume of the pyramid is $\frac{s^3}{6}$s36 and the volume of the prism with base equal to the base of the pyramid and height equal to the height of the pyramid is $\frac{s^3}{2}$s32
| $\text{Volume of Pyramid }$Volume of Pyramid | $=$= | $\text{something }\times\text{Volume of Prism }$something ×Volume of Prism | |
| $V_{Py}$VPy | $=$= | $\text{something }\times V_{Pr}$something ×VPr | Abbreviate |
| $\frac{s^3}{6}$s36 | $=$= | $\text{something }\times\frac{s^3}{2}$something ×s32 | Substitute with rules needed |
| $\frac{2s^3}{6}$2s36 | $=$= | $\text{something }\times s^3$something ×s3 | Multiply both sides by $2$2 |
| $\frac{s^3}{3s^3}$s33s3 | $=$= | $\text{something }$something | Divide both sides by $s^3$s3 |
Where the factor 'something ', we are looking for is $\frac{1}{3}$13.
So what we can see here is that the volume of the pyramid is $\frac{1}{3}$13 of the volume of the prism with base and height of the pyramid.
Of course, this is just a simple, specific example so we can get the idea of what is happening.
$\text{Volume of a pyramid }=\frac{1}{3}\times\text{area base }\times\text{height }$Volume of a pyramid =13×area base ×height
$V=\frac{1}{3}Ah$V=13Ah
Practice questions
QUESTION 4
Find the volume of the square pyramid shown.
QUESTION 5
A small square pyramid of height $4$4 cm was removed from the top of a large square pyramid of height $8$8 cm forming the solid shown. Find the exact volume of the solid.
Give your answer in exact form.
Cones
The volume of a cone has the same relationship to a cylinder as a pyramid has with a prism. So,
$\text{Volume of a right cone }=\frac{1}{3}\times\text{Area of base }\times\text{height of cylinder}$Volume of a right cone =13×Area of base ×height of cylinder
$V=\frac{1}{3}\pi r^2h$V=13πr2h
Practice questions
QUESTION 6
Find the volume of the cone shown.
Round your answer to two decimal places.
QUESTION 7
A small square pyramid of height $4$4 cm was removed from the top of a large square pyramid of height $8$8 cm forming the solid shown. Find the exact volume of the solid.
Give your answer in exact form.
Spheres
The volume of a sphere with radius $r$r can be calculated using the following formula:
$\text{Volume of a sphere }=\frac{4}{3}\pi r^3$Volume of a sphere =43πr3
Practice questions
QUESTION 8
Find the volume of the sphere shown.
Round your answer to two decimal places.
QUESTION 9
A sphere has a radius $r$r cm long and a volume of $\frac{343\pi}{3}$343π3 cm3. Find the radius of the sphere.
Round your answer to two decimal places.
Enter each line of working as an equation.