1. You can use Integration to find areas and volumes
𝑦= 𝑥
This would be the solid formed x x a b
You already know how to find the area under a curve by Integration
In this section you will learn how to find the volume of any solid created in this way. It also involves Integration!
Imagine we rotated the area shaded around the x-axis
 What would be the shape of the solid formed? 6I

2. Integration
You can use Integration to find areas and volumes y y dx dx y y x x a b a b
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) 6J

3. Integration 6J 𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥 𝑐𝑜𝑠2𝐴≡1−2𝑠𝑖 𝑛 2 𝐴
𝑐𝑜𝑠4𝐴≡1−2𝑠𝑖 𝑛 2 2𝐴
2𝑠𝑖 𝑛 2 2𝐴≡1−𝑐𝑜𝑠4𝐴
𝑠𝑖 𝑛 2 2𝐴≡ 1 2 (1−𝑐𝑜𝑠4𝐴)
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥
You can use Integration to find areas and volumes The volume of revolution of a solid rotated 2π radians around the x-axis between x = a and x = b is given by: 1) The region R is bounded by the curve y = sin2x, the x-axis and the vertical lines x = 0 and x = π/2. Find the volume of the solid formed when the region is rotated 2π radians about the x-axis.
Sub in a, b and y
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 0 𝜋 2 (𝑠𝑖𝑛2𝑥) 2 𝑑𝑥
Square the bracket
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 0 𝜋 2 𝑠𝑖 𝑛 2 2𝑥 𝑑𝑥
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥
Using the identity Cos2A = 1 – 2sin2A, replace sin22x with something equivalent
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 0 𝜋 (1−𝑐𝑜𝑠4𝑥) 𝑑𝑥
The 1/2 can be put outside the integral
𝑉𝑜𝑙𝑢𝑚𝑒= 1 2 𝜋 0 𝜋 2 (1−𝑐𝑜𝑠4𝑥) 𝑑𝑥
Integrate and use a square bracket with the limits
𝑉𝑜𝑙𝑢𝑚𝑒= 1 2 𝜋 𝑥− 1 4 𝑠𝑖𝑛4𝑥 0 𝜋 2
Sub in the two limits
𝑉𝑜𝑙𝑢𝑚𝑒= 1 2 𝜋 𝜋 2 − 1 4 𝑠𝑖𝑛2𝜋 − 0− 1 4 𝑠𝑖𝑛0
And finally we have the volume!
𝑉𝑜𝑙𝑢𝑚𝑒= 𝜋 2 4 6J

4. Integration 6J 𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥 𝑑𝑡 𝑑𝑡 𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥
Integration
You can use Integration to find areas and volumes The volume of revolution can also be calculated when x and y are given parametrically. In this case you must also include dx/dt in the integral. You will also need to change limits so they are in terms of t rather than x!
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥 𝑑𝑡 𝑑𝑡 6J

5. Integration 6J 𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥 𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥 𝑑𝑡 𝑑𝑡
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥
Integration
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥 𝑑𝑡 𝑑𝑡
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥 𝑑𝑡 𝑑𝑡
You can use Integration to find areas and volumes The volume of revolution can also be calculated when x and y are given parametrically. In this case you must also include dx/dt in the integral. The curve C has parametric equations: Where t ≥ 0. The region R is bounded by C, the x-axis and the lines x = 0 and x = 2. Find the volume of the solid formed when R is rotated 2π radians about the x-axis.
Replace y, and calculate dx/dt
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑡 2
1+2𝑡 𝑑𝑡
Square the bracket and combine them
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 1+2𝑡 (1+𝑡) 𝑑𝑡
We need to use partial fractions here
𝑦= 1 1+𝑡
1+2𝑡 (1+𝑡) 2 ≡
𝐴 (1+𝑡) 2
+ 𝐵 1+𝑡
𝑥=𝑡(1+𝑡)
𝑥=𝑡+ 𝑡 2
Combine with a common denominator
𝑑𝑥 𝑑𝑡 =1+2𝑡
1+2𝑡 (1+𝑡) 2 ≡
𝐴+𝐵(1+𝑡) (1+𝑡) 2
1+2𝑡=𝐴+𝐵(1+𝑡)
Sub in values to find A and B
t = -1
−1=𝐴
t = 0
1=𝐴+𝐵
We know A = -1 from before
2=𝐵
1+2𝑡 (1+𝑡) 2 ≡
2 1+𝑡
−1 (1+𝑡) 2
− 1 (1+𝑡) 2 𝑡 6J

6. Integration 6J 𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥 𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥 𝑑𝑡 𝑑𝑡
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥
Integration
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥 𝑑𝑡 𝑑𝑡
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥 𝑑𝑡 𝑑𝑡
You can use Integration to find areas and volumes The volume of revolution can also be calculated when x and y are given parametrically. In this case you must also include dx/dt in the integral. The curve C has parametric equations: Where t ≥ 0. The region R is bounded by C, the x-axis and the lines x = 0 and x = 2. Find the volume of the solid formed when R is rotated 2π radians about the x-axis.
Replace y, and calculate dx/dt
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑡 2
1+2𝑡 𝑑𝑡
Square the bracket and combine them
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 1+2𝑡 (1+𝑡) 𝑑𝑡
We need to use partial fractions here
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 2 1+𝑡 − 1 (1+𝑡) 2
𝑦= 1 1+𝑡
𝑥=𝑡(1+𝑡)
𝑥=𝑡+ 𝑡 2
We also need to calculate the limits for t rather than x
 Sub in the x limits and solve for t
𝑑𝑥 𝑑𝑡 =1+2𝑡
𝑥=𝑡(1+𝑡)
x = 0
x = 2
0=𝑡(1+𝑡)
2=𝑡(1+𝑡)
𝑡=0 𝑜𝑟 −1
0= 𝑡 2 +𝑡−2
0=(𝑡+2)(𝑡−1)
𝑡=0
𝑡=−2 𝑜𝑟 1
𝑡=1 6J

7. Integration 6J 𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥 𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥 𝑑𝑡 𝑑𝑡
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥
Integration
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥 𝑑𝑡 𝑑𝑡
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥 𝑑𝑡 𝑑𝑡
You can use Integration to find areas and volumes The volume of revolution can also be calculated when x and y are given parametrically. In this case you must also include dx/dt in the integral. The curve C has parametric equations: Where t ≥ 0. The region R is bounded by C, the x-axis and the lines x = 0 and x = 2. Find the volume of the solid formed when R is rotated 2π radians about the x-axis.
Replace y, and calculate dx/dt
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑡 2
1+2𝑡 𝑑𝑡
Square the bracket and combine them
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 1+2𝑡 (1+𝑡) 𝑑𝑡
We need to use partial fractions here
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 2 1+𝑡 − 1 (1+𝑡) 2
𝑦= 1 1+𝑡
𝑥=𝑡(1+𝑡)
𝑥=𝑡+ 𝑡 2
Use t-limits
𝑑𝑥 𝑑𝑡 =1+2𝑡
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑡 − 1 (1+𝑡) 2 6J

8. Integration 6J 𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥 𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥 𝑑𝑡 𝑑𝑡
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥
Integration
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑎 𝑏 𝑦 2 𝑑𝑥 𝑑𝑡 𝑑𝑡
𝑉𝑜𝑙𝑢𝑚𝑒=𝜋 𝑡 − 1 (1+𝑡) 2
You can use Integration to find areas and volumes The volume of revolution can also be calculated when x and y are given parametrically. In this case you must also include dx/dt in the integral. The curve C has parametric equations: Where t ≥ 0. The region R is bounded by C, the x-axis and the lines x = 0 and x = 2. Find the volume of the solid formed when R is rotated 2π radians about the x-axis.
Integrate and write as a square bracket
=𝜋 2 ln 1+𝑡 𝑡 0 1
Sub in limits separately
=𝜋 2𝑙𝑛 −(2𝑙𝑛1+1)
Simplify/Calculate
𝑦= 1 1+𝑡
𝑥=𝑡(1+𝑡)
=𝜋 2𝑙𝑛2− 1 2
𝑥=𝑡+ 𝑡 2
𝑑𝑥 𝑑𝑡 =1+2𝑡 6J