Lecture 7 [ Area Between Curves ]

Applications of the definite integral

Area of Region Between curves
(1) If 𝑓(𝑥) ≥ 0 on [ 𝑎, 𝑏 ], then the area under
the curve 𝑦 = 𝑓(𝑥) over [ 𝑎, 𝑏 ] is,

𝑏
𝐴 = ∫ 𝑓(𝑥 ) 𝑑𝑥
𝑎

(2) If 𝑓(𝑥) and 𝑔(𝑥) are continuous with 𝑓(𝑥) ≥ 𝑔(𝑥) on [ 𝑎, 𝑏 ], then the
area of the region between the curves 𝑦 = 𝑓(𝑥) and 𝑦 = 𝑔(𝑥) from 𝑎 to
𝑏 is
𝑏
𝐴 = ∫ [ 𝑓(𝑥 ) − 𝑔(𝑥 ) ] 𝑑𝑥
𝑎

Example: Find the area bounded by the graph of 𝒚 = 𝟑 − 𝒙 and 𝒚 = 𝒙𝟐 − 𝟗

Solution
To get points of intersection we solve

3 − 𝑥 = 𝑥2 − 9 ⟹ (𝑥 + 4)(𝑥 − 3) = 0

Hence 𝑥 = −4 and 𝑥 = 3.

3
𝐴 = ∫ [(3 − 𝑥) − (𝑥 2 − 9)] 𝑑𝑥
−4

3
𝑥2 𝑥3 343
= [ 12𝑥 − − ] = .
2 2 −4 6
Example: Find the area of the region bounded by the graphs of 𝒚 = 𝐜𝐨𝐬 𝒙 ,
𝒚 = 𝒙𝟐 + 𝟐 for 𝟎 ≤ 𝒙 ≤ 𝟐.
Solution

2
𝐴 = ∫ [ 𝑦𝑡𝑜𝑝 − 𝑦𝑏𝑜𝑡𝑡𝑜𝑚 ] 𝑑𝑥
0

2
𝐴 = ∫ [ (𝑥 2 + 2) − cos 𝑥 ] 𝑑𝑥
0

2
𝑥3 20
𝐴=[ + 2𝑥 − sin 𝑥 ] = − sin(2).
3 0 3

Sometimes, the upper and lower boundary is not defined by a single rule as
in the following example

Example: Find the area bounded by the graphs of 𝒚 = 𝒙𝟐 and 𝒚 = 𝟐 − 𝒙𝟐 for

𝟎 ≤ 𝒙 ≤ 𝟐.
Solution

To find the points of intersection we solve,

𝑥2 = 2 − 𝑥2 ⟹ 𝑥2 = 1

⟹ 𝑥 = ±1

1 2 2
⟹ 𝐴 = ∫0 [(2 − 𝑥2 ) − 𝑥2 ]𝑑𝑥 + ∫1 [𝑥 − (2 − 𝑥2 )] 𝑑𝑥
1 2
2𝑥 3 2𝑥 3
= [ 2𝑥 − ] + [ − 2𝑥 ] = 4 .
3 0 3 1
 Some regions are best treated by regarding 𝒙 as a function of 𝒚. If a

region is bounded by curves with equations

𝒙 = 𝒇(𝒚), 𝒙 = 𝒈(𝒚), 𝒚 = 𝒄 and 𝒚 = 𝒅

where 𝑓(𝑦) ≥ 𝑔(𝑦) on 𝑐≤𝑦≤𝑑

then this area is given by

𝑑
𝐴 = ∫ [ 𝑓(𝑦) − 𝑔(𝑦) ] 𝑑𝑦
𝑐

Example: Find the area enclosed by the line 𝒚 = 𝒙 − 𝟏 and the

parabola 𝒚𝟐 = 𝟐𝒙 + 𝟔
Solution
To get the intersection points we solve

𝑦2
𝑥 = 𝑦 + 1 𝑎𝑛𝑑 𝑥= −3
2

→ 𝑦 2 − 2𝑦 − 8 = 0

→ (𝑦 − 4)(𝑦 + 2) = 0

→ 𝑦 = −2 , 𝑦=4

4
𝐴 = ∫ [ 𝑥𝑟𝑖𝑔ℎ𝑡 − 𝑥𝑙𝑒𝑓𝑡 ] 𝑑𝑦
−2

4
𝑦2
𝐴 = ∫ [ (𝑦 + 1) − ( − 3) ] 𝑑𝑦
−2 2

𝑦2 𝑦3 4
𝐴=[ + 4𝑦 − ] = 18.
2 6 −2
 Note: In the last example we could found the area by integrating with respect
to 𝑥 instead of 𝑦, but the calculation is much more involved. It would have meant
splitting the region in two and computing the areas labelled 𝐴1 and 𝐴2 .

𝐴 = 𝐴1 + 𝐴2
−1 5
𝐴 = ∫ [√2𝑥 + 6 − ( −√2𝑥 + 6) ] 𝑑𝑥 + ∫ [√2𝑥 + 6 − (𝑥 − 1) ] 𝑑𝑥 = 18
−3 −1

Example: Find the area of the region in the 1st quadrant that is bounded
above by 𝒚 = √𝒙 and below by the 𝒙 − 𝒂𝒙𝒊𝒔 and the line 𝒚 = 𝒙 − 𝟐.

Solution
To get the intersection points we solve

𝑥 = 𝑦2 and 𝑥 =𝑦+2
→ 𝑦2 − 𝑦 − 2 = 0

→ (𝑦 + 1)(𝑦 − 2) = 0

→ 𝑦 = −1 , 𝑦=2

2
𝐴 = ∫0 [ 𝑥𝑟𝑖𝑔ℎ𝑡 − 𝑥𝑙𝑒𝑓𝑡 ] 𝑑𝑦

2
𝐴 = ∫ [ (𝑦 + 2) − ( 𝑦 2 ) ] 𝑑𝑦
0

𝑦2 𝑦3 2 8 10
𝐴= [ + 2𝑦 − ] =2+4− = .
2 3 0 3 3
Example: Find the area of the region bounded by the curves
𝝅
𝒚 = 𝐬𝐢𝐧 𝒙 , 𝒚 = 𝐜𝐨𝐬 𝒙 , 𝒙 = 𝟎 𝐚𝐧𝐝 𝒙 = .
𝟐
Solution
To get the intersection points we solve
𝜋
sin 𝑥 = cos 𝑥 0≤𝑥≤
2

⟹ tan 𝑥 = 1
𝜋
⟹ 𝑥=
4

𝜋
Observe that cos 𝑥 ≥ sin 𝑥 when 0 ≤ 𝑥 ≤ ,
4
𝜋 𝜋
but sin 𝑥 ≥ cos 𝑥 when ≤𝑥≤
4 2

Therefore the required area is

𝐴 = 𝐴1 + 𝐴2
𝜋
𝜋
4 1 1
𝐴1 = ∫ (cos 𝑥 − sin 𝑥) 𝑑𝑥 = [sin 𝑥 + cos 𝑥 ]04 =( + ) − (0 + 1)
0 √2 √2
2
= −1
√2

𝜋
𝜋
2 1 1
𝐴2 = ∫ (sin 𝑥 − cos 𝑥) 𝑑𝑥 = [− cos 𝑥 − sin 𝑥 ]𝜋2 = (0 − 1) − (− − )
𝜋 √2 √2
4
4

2
𝐴2 = −1 +
√2
2 2
𝐴 = 𝐴1 + 𝐴2 = ( − 1 ) + ( −1 + ) = 2√2 − 2.
√2 √2
Assignment (7)

(1) Use calculus to find the area of the triangle with vertices (2, 0), (0,2), (−1, 1).

(2) Sketch and find the area bounded by the graphs of.

(a) 𝑦 = 4 − 𝑥2 , 𝑦 = −𝑥 + 2 .

(b) 𝑦 = 𝑒𝑥 , 𝑦 = 2𝑒 −𝑥 + 1 , 𝑥 = 0.

(c) 𝑦 = 𝑥2 , 𝑦=𝑥, 𝑥 = 0, 𝑥 = 2.

(d) 𝑦 = cos 𝑥 , 𝑦 = 2 − cos 𝑥 , 𝑥 = 0, 𝑥 = 2𝜋.

(e) 𝑥 = 𝑦2 , 𝑦 = 𝑥 − 2.
4
(f) 𝑦 = 2−𝑥 , 𝑦 = 4, 𝑥 = 0.

𝜋
(g) 𝑦 = sin 𝑥 , 𝑦 = cos 𝑥 , 𝑥 =0, 𝑥 = 2.

(h) 𝑦 = 𝑥2 , 𝑦 = 𝑥3 , 𝑥 = 0, 𝑥 = 2.

(i) 𝑦 = 𝑥 2 + 2𝑥 , 𝑦 = −𝑥 + 4 , 𝑥 = −2 , 𝑥 = −4.
𝜋 𝜋
(j) 𝑦 = tan 𝑥 , 𝑦 = 0, 𝑥 = −4 , 𝑥 = 3.

(k) 𝑦 = √𝑥 − 1 , 𝑥 = 0, 𝑥 = 4.

(l) 𝑦 = − 𝑥 2 − 2𝑥 , 𝑦 = 𝑥 2 − 4.
Lecture 8 [ Volume of Revolution ]

Applications of The Definite Integral

The volume of a solid of revolution

If a region in the plane is revolved about a line, the resulting solid is a solid of
revolution, and the line is called the axis of revolution.

I) The Disk Method

As long as the rotational solid resulting from your graph has no hollow space in
it, you can use the disk method to calculate its volume.
Example: Let R be the region bounded by the curve 𝒚 = 𝒇(𝒙) = (𝒙 +
𝟏)𝟐 , the 𝒙 − 𝐚𝐱𝐢𝐬, and the lines 𝒙 = 𝟎 𝐚𝐧𝐝 𝒙 = 𝟐. Find the volume of the
solid of revolution obtained by revolving 𝑹 about the 𝒙 − 𝐚𝐱𝐢𝐬.

Solution

2
Volume = 𝜋 ∫0 ((𝑥 + 1)2 )2 𝑑𝑥
2
= 𝜋 ∫0 (𝑥 + 1)4 𝑑𝑥
2
( 𝑥 +1 )5 242 𝜋
= 𝜋 [ ] = .
5 0 5

Revolving About a Line That Is Not a Coordinate Axis

Example: Find the volume of the solid formed by revolving the region
bounded by 𝒇(𝒙) = 𝟐 − 𝒙𝟐 𝒂𝒏𝒅 𝒈(𝒙) = 𝟏 about the line 𝒚 = 𝟏.

Solution
Solve 2 − 𝑥 2 = 1 to determine that the limits

limits of integration are ±1, and

𝑅(𝑥) = ( 2 − 𝑥 2 ) − 1 = 1 − 𝑥 2

The volume is given by

1
V = 𝜋 ∫−1 (1 − 𝑥 2 )2 𝑑𝑥

16 𝜋
= .
5
Example: Find the volume of the solid generated by revolving the region
between the parabola 𝒙 = 𝒚𝟐 + 𝟏 𝐚𝐧𝐝 𝐭𝐡𝐞 𝐥𝐢𝐧𝐞 𝒙 = 𝟑 about the line.
Solution
𝑅(𝑦) = 3 − ( 𝑦 2 + 1 ) = 2 − 𝑦 2
The volume is given by
√2
V = 𝜋 ∫−√2 (2 − 𝑦 2 )2 𝑑𝑦

√2
V = 𝜋 ∫−√2 ( 4 − 4𝑦 2 + 𝑦 4 ) 𝑑𝑦

4𝑦 3 𝑦5 √2 64√2 𝜋
V = 𝜋 [ 4𝑦 − + ] = .
3 5 −√2 15

II) The Washer Method

If a solid of revolution has any hollow spots in it, you’ll have to use a
modified version of the disk method, called the washer method. It gets
its name from the fact that a thin slap of the solid resembles a circular
washer of outer radius 𝑅 and inner radius 𝑟.

Horizontal axis of revolution

Horizontal axis of
revolution

𝑏
Volume = 𝜋 ∫𝑎 ( 𝑅 2 (𝑥 ) − 𝑟 2 (𝑥 ) ) 𝑑𝑥
Vertical axis of revolution Vertical axis of revolution

𝑑
Volume = 𝜋 ∫𝑐 ( 𝑅 2 (𝑦) − 𝑟 2 (𝑦) ) 𝑑𝑦

Integrating with respect to 𝒚, two integral case

Example: Find the volume of the solid formed by revolving the region
bounded by the graphs of 𝒚 = 𝒙𝟐 + 𝟏, 𝒚 = 𝟎, 𝒙 = 𝟎, 𝐚𝐧𝐝 𝒙 = 𝟏 about the
𝒚 − 𝐚𝐱𝐢𝐬.

Solution
For the region shown , the outer radius is 𝑅(𝑦) = 1.

There is however, no convenient formula that represents

the inner radius. When 0 ≤ 𝑦 ≤ 1, 𝑟(𝑦) = 0, but when

1 ≤ 𝑦 ≤ 2, 𝑟(y) is determined by the equation 𝑦 = 𝑥 2 + 1,

which implies that 𝑟(𝑦) = √ 𝑦 − 1.

0, 0≤𝑦≤1
𝑟(𝑦) = {
√𝑦−1, 1≤𝑦≤2

The volume is given by,

1 2
2
𝑉 = 𝜋 ∫ (12 − 02 ) 𝑑𝑦 + 𝜋 ∫ ( 12 − (√ 𝑦 − 1 ) ) 𝑑𝑦
0 1

1 2
3𝜋
𝑉 = 𝜋∫ 𝑑𝑦 + 𝜋 ∫ ( 2 − 𝑦) 𝑑𝑦 = .
0 1 2
1
Note that the first integral 𝜋 ∫0 𝑑𝑦 represents the volume of a right circular cylinder of radius 1
and height 1. This portion of the volume could have been determined without using calculus.
Revolving About a Line That Is Not a Coordinate Axis
Example: Find the volume of the solid generated when the region under the
curve
𝒚 = 𝒙𝟐 over the interval [𝟎, 𝟐] is rotated about the line 𝒚 = −𝟏.

Solution
𝑅(𝑥) = 𝑥 2 − (−1) = 𝑥 2 + 1

𝑟(𝑥) = 0 − (−1) = 1

The volume is given by,

2
𝑉 = 𝜋 ∫ [ (𝑥 2 + 1)2 − (1)2 ] 𝑑𝑥
0
2 2
4 2
𝑥5 2𝑥 3 176
= 𝜋 ∫ ( 𝑥 + 2𝑥 ) 𝑑𝑥 = 𝜋 [ + ] = 𝜋.
0 5 3 0 15

Example: Find the volume of the solid obtained by rotating the region
bounded by 𝒚 = 𝒙𝟑 , 𝒚 = 𝟎 , 𝒙 = 𝟏 about the line 𝒙 = 𝟐.
Solution
3
𝑅(𝑦) = 2 − √𝑦

𝑟(𝑦) = 2 − 1 = 1

The volume is given by,

1
𝑉 = 𝜋 ∫0 [ ( 2 − 3√𝑦 )2 − (1)2 ] 𝑑𝑦

1
3
𝑉 = 𝜋 ∫ [ 4 − 4 3√𝑦 − √𝑦 2 − 1 ] 𝑑𝑦
0

1
3
𝑉 = 𝜋 ∫ [ 4 − 4 3√𝑦 − √𝑦 2 − 1 ] 𝑑𝑦
0

3 5⁄3 1 3
𝑉 = 𝜋 [ 3𝑦 − 3𝑦 4⁄3 + 𝑦 ] = 𝜋.
5 0 5
 1 𝜋
19. 𝑉 = 𝜋 ∫0 ( 𝑥 3 )2 𝑑𝑥 = .
7

1 𝜋
20. 𝑉 = 𝜋 ∫0 ( 1 − 3√𝑦 )2 𝑑𝑦 = 10 .

1 2𝜋
21. 𝑉 = 𝜋 ∫0 [ (1)2 − ( 3√𝑦 )2 ] 𝑑𝑦 = .
5

1 5𝜋
22. 𝑉 = 𝜋 ∫0 [ (1)2 − ( 1 − 𝑥 3 )2 ] 𝑑𝑥 = .
14

1 𝜋
23. 𝑉 = 𝜋 ∫0 [ (1)2 − ( √𝑥 )2 ] 𝑑𝑥 = .
2

1 7𝜋
24. 𝑉 = 𝜋 ∫0 [ (1)2 − ( 1 − 𝑦 2 )2 ] 𝑑𝑦 = .
15

1 𝜋
25. 𝑉 = 𝜋 ∫0 ( 𝑦 2 )2 𝑑𝑦 = .
5

1 2 𝜋
26. 𝑉 = 𝜋 ∫0 ( 1 − √𝑥 ) 𝑑𝑥 = .
6

1 2 5𝜋
27. 𝑉 = 𝜋 ∫0 [ (√𝑥 ) − (𝑥 3 )2 ] 𝑑𝑥 = .
14

1 13𝜋
28. 𝑉 = 𝜋 ∫0 [ (1 − 𝑦 2 )2 − ( 1 − 3√𝑦 )2 ] 𝑑𝑦 = .
30

1 2𝜋
29. 𝑉 = 𝜋 ∫0 [ ( 3√𝑦 )2 − ( 𝑦 2 )2 ] 𝑑𝑦 = .
5

1 2 10𝜋
30. 𝑉 = 𝜋 ∫0 [ ( 1 − 𝑥 3 )2 − ( 1 − √𝑥 ) ] 𝑑𝑥 = .
21
Assignment (8)
Use the disk or washer method to find the volume of revolution that is formed by revolving the
region bounded by the graphs of the given equations about the indicated line or axis.

a) 𝑦 = 9 − 𝑥 2 , 𝑦 = 0; 𝒙 − 𝐚𝐱𝐢𝐬.

b) 𝑦 = 9 − 𝑥 2 , 𝑥 = 0 , 𝑥 = 3; 𝒙 = 𝟑.

c) 𝑦 = 9 − 𝑥 2 , 𝑥 = 0 , 𝑥 = 3; 𝒚 = 𝟏𝟐.

d) 𝑦 = (𝑥 − 2)2 , 𝑥 = 0 , 𝑦 = 0 ; 𝒚 − 𝐚𝐱𝐢𝐬.

e) 𝑦 = 𝑥 2 + 1 , 𝑦 = 𝑥+3; 𝒙 − 𝐚𝐱𝐢𝐬

f) 𝑥 + 𝑦 = 1 , 𝑦 = 𝑥 + 1, 𝑥 =2; 𝒚 − 𝐚𝐱𝐢𝐬

g) 𝑦 = 1 − 𝑥 2 , 𝑦 = 𝑥 2 − 1, 𝑥 ≥ 0, 𝒚 − 𝐚𝐱𝐢𝐬.

h) 𝑦 = 𝑥 2 + 2, 𝑦 = 4 − 𝑥 2 , 𝑥 = 0, 𝑥 = 3 ; 𝒚 = 𝟏𝟐.

i) 𝑦 = √𝑥 − 1 , 𝑥 = 5 ; 𝑦 = 0; 𝒙 = 𝟓.
𝜋
j) 𝑦 = tan 𝑥 , 𝑦 = 0, 𝑥 = 4 ; 𝒙 − 𝐚𝐱𝐢𝐬.

k) 𝑦 = tan 𝑥 , 𝑦 = sec 𝑥 , 𝑥 = 0, 𝑥 = 1; 𝒙 − 𝐚𝐱𝐢𝐬.

9
l) 𝑦 = , 𝑦 = 10 − 𝑥 2 ; 𝒚 = 𝟏𝟐.
𝑥2

m) 𝑦 = 𝑒 −𝑥 , 𝑦 = 1 − 𝑒 −𝑥 , 𝑥 = 0 ; 𝒚 = 𝟒.

n) 𝑦 = cosh 𝑥 , 𝑥 = 2, 𝑥 = −2 , 𝑦 = 0; 𝒙 − 𝐚𝐱𝐢𝐬.

o) 𝑦 = ln 𝑥 , 𝑦 = 0 , 𝑥 = 0, 𝑦 = 1 ; 𝒙 = 𝟎.
1
p) 𝑦 = 𝑥 3 , 𝑦 = 2𝑥 ; 𝒙 − 𝐚𝐱𝐢𝐬
8