Volume (Edexcel IGCSE Maths)

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Volume

What is volume?

  • The volume of a 3D shape is a measure of how much space it takes up
  • You need to be able to calculate the volumes of a number of common 3D shapes

How do I find the volume of cuboids, prisms, and cylinders?

  • To find the volume of a cuboid use the formula
    Volume of a cuboid = length × width × height

Cuboid volume, IGCSE & GCSE Maths revision notes

    • You will sometimes see the terms  'depth' or 'breadth' instead of 'height' or 'width'
    • A cuboid is in fact another name for a rectangular-based prism
  • To find the volume of a prism use the formula
    Volume of a prism = area of cross-section × length

Prism volume, IGCSE & GCSE Maths revision notes

    • Note that the cross-section can be any shape, so as long as you know its area and length, you can calculate the volume of the prism
    • Or if you know the volume and length of the prism, you can calculate the cross-section area
  • To calculate the volume of a cylinder with radius, r and height, h, use the formula
    Volume of a cylinder = πr2h

Cylinder volume, IGCSE & GCSE Maths revision notes

    • Note that a cylinder is in fact a circular-based prism: its cross-section is a circle with area πr2, and its length is h

How do I find the volume of cones, & spheres?

  • To calculate the volume of a cone with base radius r and height h, use the formula
    Volume of a cone = 1/3 πr2h

Cone volume, IGCSE & GCSE Maths revision notes

    • Note that a cone is in fact a circular-based pyramid
    • As with a pyramid, to use the cone volume formula the height must be a line from the top of the cone that is perpendicular to the base
  • To calculate the volume of a sphere with radius r, use the formula
    Volume of a sphere = 4/3 πr3

Sphere Radius r, IGCSE & GCSE Maths revision notes

Examiner Tip

  • The formula for volume of a sphere or volume of a cone will be given to you in an exam question if you need it
  • You need to memorise the other volume formulae

Worked example

A sculptor has a block of marble in the shape of a cuboid, with a square base of side 35 cm and a height of 2 m.

He carves the block into a cone, with the same height as the original block and with a base diameter equal to the side length of the original square base.

What is the volume of the marble he removes from the block whilst carving the cone.

Give your answer in m3, rounded to 3 significant figures.

The volume of the removed material will be equal to the volume of the original marble minus the volume of the cone.

Find the volume of the original marble.

Convert the units of the length, width and height of the cuboid into the same units, either metres or cm.
The question asks for the answer in m3 so it makes sense to use this throughout your calculations.

Length = width = 0.35 m

Height = 2 m 

Substitute the values into the formula for the volume of a cuboid.

table row cell Volume space of space cuboid space end cell equals cell space length space cross times space width space cross times space height space end cell row blank equals cell space 0.35 space cross times space 0.35 space cross times space 2 space end cell row blank equals cell space 0.245 space m cubed end cell end table

Find the radius of the base of the cone, this will be half of the diameter.

table row cell Radius space of space base space of space cone space end cell equals cell 1 half open parentheses diameter close parentheses end cell row blank equals cell 1 half open parentheses 0.35 close parentheses end cell row blank equals cell space 0.175 space straight m end cell end table

Find the volume of the cone by substituting the radius and the height into the formula for the volume of a cone.

table row cell Volume space of space cone end cell equals cell space 1 third space cross times space area space base space cross times space height space end cell row blank equals cell 1 third cross times space pi r to the power of italic 2 space cross times space 2 space end cell row blank equals cell space 2 over 3 cross times space pi space cross times space open parentheses 0.175 close parentheses squared end cell row blank equals cell space 0.0641409... end cell end table

  

Find the volume of the marble that was removed by subtracting the volume of the cone from the volume of the cuboid.

Volume removed = 0.245 - 0.0641409 = 0.180859...

Round the answer to 3 significant figures.

Volume of removed marble = 0.181 m3 (3 s.f.)

Problem Solving with Volumes

How can I solve problems when its not a "standard" 3D shape?

  • Often the shape in a question will not be a standard cuboid, cone, sphere, etc
  • It will likely either be:
    • A prism (3D shape with the same cross-section running through it)
    • A portion or fraction of a standard shape (a hemisphere for example)
  • If the shape is a prism, recall that the volume of a prism is the cross sectional area × its length
    • The cross-sectional area may be a compound shape, such an an L-shape, or a combination of a rectangle and a triangle for example
  • If the shape is a fraction of a standard shape, consider the "full" version of the shape and then find the appropriate fraction of it
    • A hemisphere is half a sphere
    • A frustum is a truncated (chopped-off) cone or pyramid
      • The volume of a frustum will be the volume of the smaller cone or pyramid subtracted from the volume of the larger cone or pyramid

Examiner Tip

  • Before you start calculating, make a quick note of your plan to tackle the question
    • e.g. "find the area of the triangle and the rectangle, add together, times by the length"

Worked example

A doll's house is in the shape of a prism pictured below. The prism consists of a cuboid with a triangular prism on top of it. The cross section of the triangular prism is an isosceles right-angled triangle. Find the volume of the doll's house.

3-7-1-problem-solving-with-volume-1

Our strategy is to find the area of the triangle and the rectangle and add them together to find the cross-sectional area, and then multiply this by the length to find the volume

As it is an isosceles triangle, length a equals 20 space cm

We can then use Pythagoras to find length b

b equals square root of 20 squared plus 20 squared end root equals square root of 800 equals 20 square root of 2

Length c will also be 20 square root of 2

Finding the area of the triangle using Area equals 1 half cross times base cross times height

Triangle space Area space equals 1 half cross times 20 cross times 20 equals 200 space cm squared

Finding the area of the rectangle

40 cross times 20 square root of 2 equals 800 square root of 2 space cm squared

The total cross-sectional area is therefore the triangle plus the rectangle

200 space plus space 800 square root of 2 space cm squared space equals space 1 space 331.37085... space cm squared

Finding the volume of the prism by multiplying the cross-sectional area by the length

open parentheses 200 plus 800 square root of 2 close parentheses cross times 60 equals 12000 plus 48000 square root of 2 space cm cubed equals 79 space 882.25099... space cm cubed

Rounding to 3 significant figures

79 900 cm3

Worked example

The diagram shows a truncated cone (a frustum). Using the given dimensions, find the volume of the frustum.

3-7-1-problem-solving-with-volume-2

To find the volume of the frustum, find the volume of the larger cone (30 cm tall, with a radius of 20 cm), and subtract the volume of the smaller cone (15 cm tall, with a radius of 10 cm)

Formula for the volume of a cone: 1 third pi space r squared straight h

Calculate the volume of the larger cone

V subscript L equals 1 third cross times straight pi cross times 20 squared cross times 30 equals 4000 straight pi equals 12 space 566.37061... space cm cubed

Calculate the volume of the smaller cone

straight V subscript straight S equals 1 third cross times straight pi cross times 10 squared cross times 15 equals 500 straight pi equals 1 space 570.796327... space cm cubed

Find the difference

V subscript L minus V subscript S equals 4000 straight pi minus 500 straight pi equals 3500 straight pi equals 10 space 995.57429... space cm cubed

Round to 3 significant figures 

11 000 cm3 

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Amber

Author: Amber

Expertise: Maths

Amber gained a first class degree in Mathematics & Meteorology from the University of Reading before training to become a teacher. She is passionate about teaching, having spent 8 years teaching GCSE and A Level Mathematics both in the UK and internationally. Amber loves creating bright and informative resources to help students reach their potential.