1. Integration
Volumes of revolution

2. FM Volumes of revolution I
KUS objectives
BAT Find Volumes of revolution using Integration
Starter: Find these integrals
𝑥 5 𝑑𝑥
= 1 6 𝑥 6 +𝐶
( 𝑥 +4 𝑥 3 )𝑑𝑥
= 2 3 𝑥 3/2 + 𝑥 4 +𝐶
1 𝑥 2 𝑑𝑥
=− 1 𝑥 +𝐶

3. Notes 1
You can use Integration to find areas and volumes y y x a b dx x a b y dx
In the trapezium rule we thought of the area under a curve being split into trapezia.
To simplify this explanation, we will use rectangles now instead
The height of each rectangle is y at its x-coordinate
The width of each is dx, the change in x values
So the area beneath the curve is the sum of ydx (base x height)
The EXACT value is calculated by integrating y with respect to x (y dx)
For the volume of revolution, each rectangle in the area would become a ‘disc’, a cylinder
The radius of each cylinder would be equal to y
The height of each cylinder is dx, the change in x
So the volume of each cylinder would be given by πy2dx
The EXACT value is calculated by integrating y2 with respect to x, then multiplying by π. (πy2 dx)
𝑎𝑟𝑒𝑎= 𝑎 𝑏 𝑦 𝑑𝑥

4. Notes 2 𝑎𝑟𝑒𝑎= 𝑎 𝑏 𝑦 𝑑𝑥 𝑣𝑜𝑙𝑢𝑚𝑒 = 𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥x y a b y x
This would be the solid formed
𝑎𝑟𝑒𝑎= 𝑎 𝑏 𝑦 𝑑𝑥
𝑣𝑜𝑙𝑢𝑚𝑒 = 𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥
Imagine we rotated the area shaded around the x-axis
What would be the shape of the solid formed?

5. 3𝑥− 𝑥 2 =2 solves to give points 1, 2 𝑎𝑛𝑑 2, 2
WB A The region R is bounded by the curve 𝑦 =2𝑥+1 and the vertical lines x=1 and x=3
Find the volume of the solid formed when the region is rotated 2π radians about the x-axis.
3𝑥− 𝑥 2 =2 solves to give points 1, 2 𝑎𝑛𝑑 2, 2
𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 𝑥 𝑑𝑥
𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 𝑥 2 +4𝑥+1 𝑑𝑥
= 158𝜋 3
See Geogebra workbookA1

6. WB A The region R is bounded by the curve 𝑦 = 1 𝑥 , the x-axis and the vertical lines x = 1 and x =2. Find the volume of the solid formed when the region is rotated 2π radians about the x-axis. Write your answer as a multiple of π
𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 𝑥 − 𝑑𝑥
𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 𝑥 −2 𝑑𝑥
= 𝜋 −𝑥 −
= 𝜋 2

7. WB A The region R is bounded by the curve 𝑦 =𝑥 𝑥 , the x-axis and the vertical lines x = 0 and x =2. Find the volume of the solid formed when the region is rotated 2π radians about the x-axis. Write your answer as a multiple of π
𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 𝑥 3/ 𝑑𝑥
𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 𝑥 3 𝑑𝑥
= 𝜋 𝑥
=4𝜋

8. WB A The region R is bounded by the curve 𝑦 = 3− 𝑥 3 , the x-axis and the vertical lines x = -1 and x = -3
Find the volume of the solid formed when the region is rotated 2π radians about the x-axis. Give your answer as an exact multiple of π
𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 −3 − − 𝑥 𝑑𝑥
𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 −3 −1 3− 𝑥 3 𝑑𝑥
= 𝜋 3𝑥 − 1 4 𝑥 4 −1 −3
= 𝜋 −3− −𝜋 −9− = 𝜋

9. 𝑦 =𝑥 1−𝑥 , so 𝑦 2 = 𝑥 2 (1−𝑥) 2 = 𝑥 2 −2 𝑥 3 + 𝑥 4
WB A The region R is bounded by the curve 𝑦 =𝑥(1−𝑥), the x-axis and the vertical lines x = 0 and x =1
Find the volume of the solid formed when the region is rotated 2π radians about the x-axis.
𝑦 =𝑥 1−𝑥 , so 𝑦 2 = 𝑥 2 (1−𝑥) 2 = 𝑥 2 −2 𝑥 3 + 𝑥 4
𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 𝑥 2 −2 𝑥 3 + 𝑥 4 𝑑𝑥
= 𝜋 𝑥 3 − 1 2 𝑥 𝑥
= 𝜋 30

10. WB A6 The region R is bounded by the curve 𝑦 =3𝑥− 𝑥 2 andand the horizontal line y=2
Find the volume of the solid formed when the region is rotated 2π radians about the x-axis.
3𝑥− 𝑥 2 =2 solves to give points 1, 2 𝑎𝑛𝑑 2, 2
𝑣𝑜𝑙𝑢𝑚𝑒=𝜋 𝑥− 𝑥 𝑑𝑥 −𝜋 𝑑𝑥
= 47𝜋 10 −4π
= 7𝜋 10

11. NOW DO Ex 5A 𝑣𝑜𝑙𝑢𝑚𝑒=2𝜋 0 𝑟 𝑥 2 − 𝑟 2 2 𝑑𝑥 𝑥 2 + 𝑦 2 = 𝑟 2 𝑦= 𝑟 2 − 𝑥 2
WB A Find the volume of the sphere formed when a circle of radius r, centred at the origin is rotated 2π radians about the x-axis.
𝑣𝑜𝑙𝑢𝑚𝑒=2𝜋 0 𝑟 𝑥 2 − 𝑟 𝑑𝑥
𝑥 2 + 𝑦 2 = 𝑟 2
𝑦= 𝑟 2 − 𝑥 2
𝑣𝑜𝑙𝑢𝑚𝑒=2𝜋 0 𝑟 𝑟 2 −𝑥 2 𝑑𝑥
=2𝜋 𝑟 2 𝑥− 1 3 𝑥 3 𝑟 0
=2𝜋 𝑟 3 − 1 3 𝑟 3 − 0 −0 = 4 3 𝜋 𝑟 3
NOW DO Ex 5A

12. One thing to improve is –
KUS objectives BAT Find Volumes of revolution using Integration
self-assess
One thing learned is –
One thing to improve is –

13. END