Download Application of Definite Integration

GROUP A – CLASS WORK PROBLEMS :
Q-1) Find the area of the region bounded by the
following curves, the X-axis and the given
lines
i) y = x4, x = 1, x = 5
Let A be the required area.
Consider the equation, y =
2
∴
A =
∫
6x + 4
2
y dx =
0
∫
6x + 4 dx
0
2
=

3 
 ( 6x + 4 ) 2 
 3


×6 
 2
 0
=
3
3
1

(6 × 2 + 4) 2 – ( 6 × 0 + 4) 2 

9

=
3
3
1

(16 ) 2 – ( 4 ) 2 

9

=
1
[64 – 8]
9
ii) y = 6x + 4, x = 0, x = 2
Ans. i)
Y
5
4
3
1 2
 4
9 
3
2
( ) ( )
=
– 22
3
2



2
1
X′
0
1
2
3
4
X
5
∴
Y′
Consider the equation, y = x4
∴
 x5 
1
5
5
A = y dx = x .dx =   = ( 5 ) – (1) 


5
5
 1
1
1
∫
=
∴
5
5
A =
∫
56
= sq.units
9
Q-2) Find the area of ∆ PQR whose vertices are
P(2, 1), Q(3, 4), and R (5, 2)
Let A be the required area.
5
A =
Ans.
4
Y
1
[3125 – 1]
5
Q(3, 4)
3124
sq.units
5
R(5, 2)
P
(2,1)
ii)
Y
X′
0
P′
Q′
R′
X
Y′
4
Equation of PQ :
3
2
(y – y1) =
1
X′
0
Y′
1
2
3
X
y1 – y2
( x – x1 )
x1 – x 2
where x1 = 2, y1 = 1 and x2 = 3, y2 = 4
(y – 1) =
1– 4
( x – 2)
2–3
Mahesh Tutorials Science
2
∴
(y – 1) = 3(x – 2)
∴
y = 3x – 6 + 1
∴
y = 3x – 5
... (i)
Equation of QR where x1 = 3, y1 = 4 and
x2 = 5, y2 = 2
=
–
1 2 5
5
x  + 7 [x ]


3
3
2
=
–
1 2
2
(5 ) – ( 3)  + 7 ( 5 – 3)

2
=
–
1
(16 ) +14
2
4–2
( x – 3)
(y – 4) =
3–5
∴
5
A( PP′′R′′R) =
( y – 4) = –1(x – 3)
∴
y = –x + 3 + 4
∴
y = –x + 7
x
= 6
1
∫  3 + 3  dx
...[from (iii)]
2
5
=
1  x2 
1
5
  + [ x ]2
3  2 2 3
=
∴
1– 2
( x – 2)
(y – 1) =
2–5
1  25 4  1
–
+ [5 – 2]
3  2 2  3
∴
1
(y – 1) = ( x – 2)
3
=
1  21 
+1 =
3  2 
∴
x 2
– +1
y=
3 3
∴
y=
... (ii)
Equation of PR where x1 = 2, x2 = 5 and
y1 = 1, y2 = 2
7
+1
2
From (iv), we get
A(∆ PQR) =
x 1
+
3 3
15
7
8
– 5 + 6 – – 1 = = 4sq.units
2
2
2
... (iii)
Q-3) Find the area of ellipse
Required area = A( PP′′Q′′Q) + A( QQ′′R′′R) –
A( PP′′R′′R)
... (iv)
x2 y2
+
= 1.
4 25
Ans.
Now,
Y
3
A( PP′′Q′′Q) =
∫ (3x – 5 )dx
Q(0,5)
...[from (i)]
2
X
3
0
 x2

– 5x 
= 3
2
2
By the symmetry of the ellipse, required area
of the ellipse is 4 times the area of the sector
OPQ.
3
(9 – 4) – 5 ( 3 – 2)
2
15
–5
=
2
Equation of ellipse is
x 2 y2
+
= 1 ... (given)
4 25
where a = 2 and b = 5
5
A( QQ′′R′′R)=
X
Y′
3
 x2 
3
= 3   – 5 [ x ]2
 2 2
=
P(2,0)
∫ ( – x + 7 ) dx
3
...[from (ii)]
∴
y2
x2
=1 –
25
4
∴

x2 
y2 = 25 1 –

4 

5
 x2

+ 7x 
= –
2

3
=
Application of Definite Integration
25
4 – x2
4
(
)
=
25  2
2
( 2) – ( x ) 
4 
Mahesh Tutorials Science
∴
y= ±
5
2
( 2)
2
3
– x2
= 4–
We only consider positive value because in
first quadrant, y > 0.
Then required area = 4(Area of sector OPQ)
2
∴
2
∫
A = 4 y.dx = 4
0
5
∫2
Area of the region bounded by the given curve
is
4
sq. units.
3
Q-5) Find area of region lying between y2 = x
4 – x 2 dx
and x2 = y
0
2
=
∴
8 4
=
3 3
4×5 x
4
 x 
4 – x 2 + sin–1   

2 2
2
 2 0
Ans.
2
x = 4y
Y


–1  2  
= 10  0 + 2sin    – [0 + 0 ]
 2 


A
B
= 10 – 2 sin–1 (1)
O
X′
π
= 20 × = 10π
2
∴
2
y = 4x
C
D
(4, 0)
X
Area of ellipse is 10π sq. units.
Y′
Q-4) Find the area of the region bounded by the
curve y = 2x – x2 and X-axis.
Ans.
For finding the points of intersection of the
two parabolas, we equate the values of y2 from
their equations.
From the equation x2 = y, y2 = x4
Y
∴ x4 = x
2
∴ x(x3 – 1) = 0
1
X
x4 – x = 0
–2 –1 0
–1
1
2
∴ x = 0, x = 1
X
When x = 0, y = 0
When x = 1, y = 1
–2
∴ the points of intersection are O(0, 0) and
A(1,1).
Y′
Equation of curve is y = 2x – x2
... (given)
Required area = area of the region OBACO
= (area of the region ODACO)
2
Then, required area, A =
∫ y.dx
0
The limit of integration is x = 0 and x = 2
because area of curve bounded is in positive
X-axis.
– (area of the region ODABO)
Now, area of the region ODACO
= area under the parabola y2 = x, i.e., y = x
1
2
∴
A =
∫ (2x – x ) dx
2
=
2
2
∫
∫
0
0
= 2 x dx –
x 2 dx
∫
0
0
=
1
 3 
 x2 
x dx = 
3 / 2


0
2
2
(1 − 0 ) =
3
3
Area of the region ODABO
1
1
2
2
3
3
= 2 × ( 2) – ( 0 )  – ( 2) – ( 0 ) 
2
2
= area under the parabola x2 = y
Application of Definite Integration
Mahesh Tutorials Science
4
1
=
∫
0
4
1
x3 
x 2 dx =  
 3  0
=
∫
–2
3x 2 3x
–
– 6 dx
4
2
4
=
1
1
(1 − 0 ) =
3
3
∴ required area =
2 1 1
− = sq unit.
3 3 3
=
 3x 3 3x 2

–
– 6x 

4
 12
 –2
=
 ( 4 )3 3 ( 4 )2


–
– 6 ( 4) –
4
 4

Q-6) Find the area cut off from the parabola
 ( –2)3 3 ( –2)2


–
– 6 ( –2) 
4
 4

4y = 3x2 with the line 2y = 3x + 12.
Ans.
Y
2
4y = 3x
=
2y = 3x + 12
B (4,12)
D (–2,3)
X′
∴
Q-7) Find the area lying above the X-axis and
included between the circle x2 + y2 = 8x
Equation of parabola is 4y = 3x2
y1 =
= 27 sq.units
Required area = 27 sq. units
X
0
Y′
∴
64
8
– 12 – 24 + + 3 – 12
4
4
3 2
x
4
and the parabola y2 = 4x.
Ans.
Y
Equation of line is 2y = 3x + 12
∴
y2 =
P (4, 4)
3
x +6
2
Let y1 = y2
∴
3x +12
3 2
x =
2
4
∴
3x2
∴
x2 – 2x – 8 = 0
∴
(x – 4)(x + 2) = 0
∴
x = 4 and x = –2
When x = –2, y =
C (4, 0)
X
Y′
Equation of circle is x2 + y2 = 8x
2
Equation of parabola is y = 4x
... (i)
... (ii)
Substituting (ii) in (i), we get
3 ( 4 ) +12
2
3 ( –2) +12
2
x2 + 4x = 8x,
= 12 and
=3
The points of intersection are D(–2, 3) and
B(4, 12).
Thus, required area, A
∴
x2 – 4x = 0,
∴
x(x – 4) = 0
∴
x=0
or
x=4
∴
y=0
or
y=4
Thus, the circle x2 + y2 = 8x with centre at
(4, 0) and radius 4 intersects the parabola
4
=
0
= 6x + 24
When x = 4, y =
∴
X
y2 = 4 x
∫y
1
y2 = 4x at O(0, 0) and P(4, 4).
– y2 dx
x2 + y2 = 8x
–2
∴
Application of Definite Integration
y=
8x – x =
(
16 16 – 8x + x 2
)
Mahesh Tutorials Science
∴
16 – ( x – 4 )
y=
5
2
The given equation of curve is
... (i)
(y –1)2 = 4(x + 1)
2
and y = 4x
∴
The given equation of line is y = x – 1
y= 2 x
... (ii)
Since the given curve is a parabola with
vertex at (–1, 1), we will shift the origin to
Required area is the difference of areas under
the curves the line x = 0 and x = 4.
the point (–1, 1).
4
Required area, A =
∫y
1
The value of the area reamains unchanged.
– y2 dx
Put x + 1 = X, y – 1 = Y
0
where y1 =
=
8x – x =
16 – ( x – 4 )
(
16 16 – 8x + x 2
∴
)
Thus, new equations of the curve are
2
Y2 = 4X i.e. Y = 2 X
And y. = 4x = 2 x
4
∴
A =
∫
x = X – 1 and y = Y + 1
4
2
16 – ( x – 4 ) dx – 2
0
∫
x dx
0
4
16
x – 4
2
x –4
16 – ( x – 4 ) +
sin –1
= 

2
2
4 0

4
 3
x2 
–2 

 3 
 2  0
=
( 0 + 0 ) – 0 + 8sin–1 (1)  – 4 ( 8 )

 3
=
 π  32
–8  –  –
 2 3
(
∴
∴
∴
∴
∴
)
and the line, Y + 1 = X – 1 – 1
i.e. Y = X – 3
... (ii)
w.r.t. the new co-ordinate system (X, Y).
Solving these equations for their point of
intersection, we get
(X – 3)3 = 4X
X2 – 6X + 9 = 4X
X2 – 10X + 9 = 0
(X – 1) (X – 9) = 0
X = 1 or X = 9 and Y = –2 or Y = 6
Q ≡ (1, –2) and P ≡ (9, 6)
Line Y = X – 3 intersects X-axis at R(3, 0).
Required area is divided into two parts,
A1 = Above X-axis
A2 = Below X-axis
Area, A1 = A(ORPO) = A(ORMPO) – A(RMPR)
0
=
∴
8

A = 4  π –  sq.units
3

=
y=x–3
P (9, 6)
2
Y = 4X
=
X=9
A2 R (3, 0) M
∫
9
2 X dX –
 3
X2
= 2
 3
 2
Ans.
Q
(1, –2)
Y′
– y2 dx
0
y = x – 1.
0
1
9
curve (y – 1)2 = 4(x + 1) and the line
X
∫y
0
Q-8) Find the area of the region bounded by the
Y
... (i)
X
∫ ( X – 3)dx
3
9

9

 X2

– 3X 
 –
 2
3

0
4
9
1
4
(1) +  – 9 – + 3  = + 2
3
2
2
 3
10
sq.units
3
∴
A2 =
∴
Required area, A = A1 + A2 = 18 +
=
10
3
64
sq.units.
3
Application of Definite Integration
Mahesh Tutorials Science
6
Q-9) Find the area of the region bounded by the
Equation (i) represents the line passing
curve y = sin x between x = 0 and x = 2π.
 1 
through (0, 1) and  – , 0  .
 2 
Ans.
Equation (ii) represents the line passing
Y
 1 
through (0, 1) and  – ,0 
 3 
P
X′
0
Equation (iii) represents the line parallel to
Y-axis and passes through (4, 0)
B X
(2 π,0)
A
(π,0)
Required area (A)
Q
Y′
=
Given function is y = sin x.
The graph of the function is as shown above
4
0
0
∫ (y of second line dx ) – ∫ (y of first line dx )
4
4
B ≡ (2π, 0).
0
0
Since the curve is symmetrical,
=
4
 3x 2

+ x  –  x 2 + x 

0
 2
0
=
  3 ( 4 )2






+
4
–
0
+
0
(
)

 – [(42 + 4) – (0 + 0)]
2


 


which intersects X-axis at A ≡ (π, 0) and
∴
4
=
4
Required Area, A = A(OPA)
π
∫ (3x +1) dx – ∫ (2x +1) dx
π
∫
∫
0
0
= 2 y dx = 2 sin x dx
π
= 2 [ – cos x ]0
= 28 – 20
= 2 [ – cos π + cos 0]
= 8 sq. units
= 2[–(–1) + 1]
Q-11) Find the area bounded by the curve
= 4 sq. units.
y = 2x – x2 and the line y = –x.
Ans.
Q-10) Find the area of the region bounded by
curves y = 2x + 1; y = 3x + 1 and x = 4
Y
Ans.
A(1,1) x = 3
Y
B
(2,0)
y = 3x + 1
X′
0
M
X
y = 2x + 1
X′
P (3,–3)
(0,1)
–1 ,0
2
–1 ,0
3
0
Y′
(4,0)
x=4
Given equations to curves are
y = 2x + 1
... (i)
y = 3x + 1
... (ii)
x=4
... (iii)
Application of Definite Integration
X
y = 2x – x
2
Y′
Given equations of curve is y = 2x – x2
and equation of line is y = – x
Now, point of intersection between two curves
can be determined as
Mahesh Tutorials Science
7
2x – x2 = – x
Ans.
2
∴
2x – x + x = 0
∴
3x – x2 = 0
∴
3x – x2 = 0
∴
x (3 – x) = 0
∴
x = 0 and x = 3
Y
4
3
2
When x = 0 then y = 0
1
When x = 3 then y = – 3
X′
Thus, points of intersection are (0, 0) and
(3, –3)
X
0
Y′
Now, area bounded by curve is
3
A=
∫ (y
1
The given equation of curve is y = 4x2
– y2 ) dx
0
2
where y1 = 2x – x and y2 = – x
∴
x2 =
y
4
∴
x=
y
1
=±
y
4
2
3
∴
A =
∫ {(2x – x ) dx – ( – x ) dx}
2
0
We have to find the area only in first quadrant.
3
3
∫
∫x
0
0
= 2 x . dx –
3
3
2
∫
dx + x . dx
0
3
4
3
 x3 
x3 
 x3 
= 2  –   +  
 2 0  3 0  2 0
= 2×
So, we only consider +
∴
A =
2
∫
y ⋅ dy
2
=
 1 +1 
1  y2 


21
+1 
 2
 2
=
3
1 2 2
× y 
2 3   2
=
3
1  32

( 4 ) – ( 2) 2 

3

1
+
[(3)2 – (0)2)]
2
4
1
1
[27] +
[9]
3
2
9
= [9 – 9] +  
2
=
4
4
1
1
[(3)2 – (0)2] –
[(3)2 – (0)3]
2
3
= [9 – 0] –
∫
1
1
y dy =
2
2
1
y
2
9
sq.units.
2
=
1 2
2
3

( )
3
2
–
3
2 2

( ) 
2
GROUP A – HOMEWORK PROBLEMS :
∴
Q-1) Find the area of the region lying in the first
quasrant and bounded by y = 4x2,
A =
1
8 – 2 2  sq. units

3
x = 0, y = 2 and y = 4.
Application of Definite Integration
Mahesh Tutorials Science
8
Q-2) Find the area of circle x2 + y2 = 25.
Substituting y = x in the equation x2 = y,
Ans.
we get x2 = x
∴
x2 – x = 0
∴
x (x – 1) = 0
∴
x = 0 or x = 1
Y
X′
When x = 0, y = 0 and when x = 1, y = 1
X
0
∴
The point of intersection are O(0,0) and
B(1, 1).
Required area =
Y′
(Area under line)–(Area under curve)
Consider equation of circle x2 + y2 = 25
∴
y = 25 – x
1
=
2
0
Put y = 0
∴
5
= 4
∫
2
25 – x dx
5
x
25
 x 
25 – x 2 +
sin–1   
= 4
2
 5 0
2
∴

25
 5 
sin–1   
= 4 0 +
2
 5 

∴
0
∫
1
xdx –
0
∫ x dx
2
0
1
=
 x2 
 x3 
–
 
 
 2 0  3 0
=
1 1
–
2 3
=
1
6
0
= 4×
∫
1
x 2dx =
1
x=±5
Area of circle
∴
∫
1
ydx –
Required area =
1
sq. units
6
Q-4) Find the area of the region bounded by the
parabola y2 =4ax and its latus rectum.
25 π
×
2 2
Ans. Given equation of parabola is y2 = 4ax
∴
Area of circle = 25π sq. units
y=2 a x
Y
Q-3) Find the area of the region bounded by the
parabola y = x2 and the line y = x in the
first quadrant.
2
y = 4ax
A
x=a
Ans.
X′
Y
0
y=x
S (a,0)
B
y = x2
B (1, 1)
X′
0
Y′
X
Y′
Required area
= Area of region OBSAO
Consider two curves x2 = y and y = x.
Application of Definite Integration
= 2(area of region OSAO)
X
Mahesh Tutorials Science
9
Thus, x = 0 and x = 1 are the limits of
a
∫
= 2 ydx
integration.
0
2
Then, let y1 = 1 – x and y2 = (1 – x)
a
∫
Required area, A
= 2 2 a xdx
0
1
= 4 a
∫ (y
=
a
∫
1
0
xdx
0
∴
1
∫
=
 x2 
x
1
 x 
1 – x 2 + sin–1    +  –


2
 1 0  2 0
2
=
 (1)2 ( 0 )2 
1
1
–1  1  

1
–
1
+
sin
+
–

   –
2
2 
 1    2
2

=
1
  1
–1
 2 sin (1)  +  – 2 

 

1 – x 2 dx +
0
∫ (1 – x ) dx
0
1
1
 3
8
a a 2 
3
 
8 2
A = a sq. units
3
Q-5) Find the area enclosed between the circle
x2 + y2 = 1 and the line x2 + y2 = 1 lying in
Ans. the first quadrant.
(0, 1)
π 1
=   –
4 2
∴
Y
X′
1
=
a
2 3 
= 4 a  x2 
 3
 0
=
+ y2 ) dx
2
Required area =
1 π

– 1 sq. units
2  2

2
x +y =1
0
(1, 0)
Q-6) Using integration, find the area of the
triangle PQR, whose vertices are at P(2,
X
5), Q(4, 7), and R(6, 2).
Y
Q (4, 7)
Y′
P (2, 5)
2
2
Given equation of circle is x + y = 1 ...(i)
Equation of line is x + y = 1
Solving (i) and (ii), we get
...(ii)
R (6, 2)
Ans.
y2 = 1 – x2
∴
y =
0
1 – x2
P′
Q′ R′
X
And y = 1 – x
( 1– x )
∴
(1 – x) =
∴
(1 – x)2 = 1 – x2
2
2
Equation of PQ =
∴
2
∴
1 + x – 2x = 1 – x
∴
1 + x2 – 2x + x2 – 1 = 0
∴
2x (x – 1) = 0
∴
x = 0 and x = 1
y – 5 5 – 7 –2
=
=
=1
x – 2 2 – 4 –2
x–2=y–5
y=x+3
Equation of QR =
∴
y=
y –7 7–2
5
=
=
x – 4 4 – 6 –2
–5
–5 x
(x – 4) + 7 =
+ 17
2
2
Application of Definite Integration
Mahesh Tutorials Science
10
Equation of PR =
Q-7) Find the area of the region bounded by
y –5 5–2
3
=
=–
x –2 2–6
4
∴
y = 4y – 20 = –3x + 6
∴
y=
curve y = sin x between x = 0 and x = 2x.
Ans.
–3
13
x+
4
2
Y
P
Required Area,
A = A( PP ′Q′) + A ( QQ′R′R) – A ( PP ′R′R)
X′
0
...(i)
Q
4
∫ ( x + 3) dx
A( PP ′ Q′ ) =
Y′
2
Given function is y = sin x.
4
 x2

+ 3x 

2

2
=
B X
(2π,0)
A
(π,0)
The graph of the function is as shown above
which intersects X-axis at A ≡ (π, 0) and
 ( 4 )2
 4

=  2 + 3 ( 4 )  –  2 + 3 ( 2) 


 
= (12 + 8) – (2 + 6)
B = (2π, 0).
Since the curve is symmetrical,
∴
Required Area, A = A(OPA)
π
= 12
π
∫
∫
0
0
= 2 y dx =2 sin x dx
6
 –5 x

A ( QQ′R′R)=  2 +17  dx


4
∫
π
= 2 [ – cos x ]0
= 2 [ – cos x + cos 0]
6
 –5x 2

+17x 
= 
 4
4
= [–45 + 102] – [–20 + 68]
= 4 sq.units
Q-8) Find the area of the region common to the
= 57 – 48
circle x2 + y2 = 9 and the parabola y2 = 8x .
= 9
Y
6
y2 = 8 x
 –3x 13 
A ( PP ′R′R)=  4 + 2  dx


2
∫
6
 –3x 2 13x 
=  8 + 2 

2
=
Ans. X ′
2

 –3 ( 36 ) 78   –3 ( 2)

+
–
+13


2   8

 8

=
Consider x2 + y2 = 9
...(i)
y2 = 8x
...(ii)
Substituting (ii) in (i), we get x2 + 8x – 9 = 0
∴
= 7 sq. units
Application of Definite Integration
X
Y′
–27
3
+ 39 + – 13
2
2
Required Area, A = 12 + 9 – 14
0
x + 9x – x – 9 = 0
x (x + 9) –1(x + 9) = 0
∴
(x – 1)(x + 9) = 0
∴
x = 1 or x = – 1
∴
Required area, A
Mahesh Tutorials Science
11
GROUP B – CLASS WORK PROBLEMS :
1
∴
= 2
∫
y1 – y 2 dx
–9
Q-1) Find the volume of the solid obtained by


9 – x 2 dx 
= 2  2 2 xdx +
0

1


1
the complete revolution of the ellipse
3
∫
∫
x2 y2
+
= 1 about
36 25
=
i) the X-axis
3
1

9
 2 3   x
 x 
9 – x 2 + sin –1   
2 2 2  x 2   + 


2
2
 3  1

3
  0 
 4 2

 3
= 2

–


 
9
 
– 0  + 0 + sin–1 (1)  
 
2
 


1
9
–1  1   
 × 2 2 + sin    
2
 3  
2
ii) the Y-axis
x 2 y2
+
=1
36 25
Ans. Equation of ellipse is
i) y2 =
25
(36 – x2)
36
Y
B
 4 2 9π
9
 1 
+
– 2 – sin–1   
= 2
4
2
 3  
 3
X′
 8 2 9π
 1 
+
– 2 2 – 9 sin–1    sq. units
= 
2
 3  
 3
0
A′
(6,0)
A
(6,0)
X
B′
Y′
Q-9) Find the area under the curve
y = (x2 + 2)2 + 2x between the line x = 0;
x = 2 and the X-axis.
Required volume of the soild obtained by
revolving the ellipse about X-axis
6
Ans. Equation of curve is
y = (x2 + 2)2 + 2x = x4 + 4x2 + 2x + 4
=
∫
π y 2dx
–6
2
Required area, A =
∫ y dx
0
6
=
π
25
∫ 36 (36 – x ) dx
2
–6
2
=
∫ (x
4
)
+ 4x 2 + 2x + 4 dx
0
=
25π
36
2
=
=
=
 x 5 4x 3

+
+ x 2 + 4x 

3
5
0
 32 32

 5 + 3 + 4 + 8  – [0]


=
50π
36
6
∫ (36 – x ) dx
2
–6
6
∫ (36 – x ) dx
2
0
6
=
50π 
x3 
36x –

36 
3 0
=

50π 
63
36
6
–
– (0 – 0)
(
)

36 
3

=
50 π
[36(6) – (6)2.(2)]
36
=
50 π
(144) =
36
436
sq. units
15
200π cubic units
Application of Definite Integration
Mahesh Tutorials Science
12
36
(25 – y2)
25
ii) x2 =
5
Required volume
=
∫
π y 2dx
4
Y
5
B (0,5)
=
∫
π 4x dx
4
X′
0
A′
A
X
=
π 2x 2 
5
4
2
= π [2(5) – 2(4)2]
= 18π cubic units
Y′
ii) Equation of curve is xy = 1
Required volume of the solid obtained by
revolving the ellipse about Y-axis
i.e. x–1 = y
x = –1 to x = –2
5
∫
2
= π x dy
x = –2 x = –1
Y
–5
5
=
π
36

2
2 
∵ x = 25 25 – y 


36
∫ 25 (25 – y ) dy
(
2
–5
=
36π
( 2)
25
)
X′
–2
–1
5
∫ (25 – y ) dy
2
Y′
0
5
=
=
X
0
2 ( 36 ) π 
y3 
25y –

25 
3 0
–1
Required volume
=
∫
2
π y –1
dx
–2
2 ( 36 ) π 

53
– ( 0 – 0 )
25 ( 5 ) –
25 
3

–1
=
∫
π x –2dx
–2
2 ( 36 ) π  3  2  
5

25   3  
–1
=
 –1 
π 
 x  –2
Q-2) Find the volume of the solid obtained by
=
1 
π 
2
revolving about the X-axis, the region
bounded by curves.
=
π
cubic units
2
=
= 240π cubic units
i) y2 = 4x and the lines x = 4 and x = 5.
ii) xy = 1 and the lines x = –1 and x = –2
Ans. i) Equation of curve is y2 = 4x, x = 4 to x = 5
Y
X′
x=4
0
x=5
X
Y′
Application of Definite Integration
Mahesh Tutorials Science
13
Q-3) The area enclosed between the parabola
2
π
2
45 – 0 
= 2π 4 – 0  –

80 
2
y = 4x and x = 4y is revolved about Xaxis. Find the volume of solid so generated.
= 32π –
Ans.
2

= 32π 1 – 
5


x2 = 4y
Y
A
y2 = 4x
X′
0
B
π
× 1024 = 32π 1 – 32 
80
80 

= 32π ×
X
=
3
5
96π
cubic units
5
Q-4) The area bounded by the curve
y2 = x2(1– x2) between x = 0 and x = 1 is
rotated about X-axis. Obtain the volume
Y′
Given curves are
of solid so generated.
2
x = 4y
... (i)
y2 = 4x
... (ii)
Ans. Equation of curve is y2 = x2(1 – x2)
x = 0 to x = 1
To find their point of intersection, putting
y=
Required volume,
x2
in equation (ii), we get
4
1
1
∫
∫ (x
0
0
V = π y 2 dx = π
2
)
– x 4 dx
2
 x2 

 = 4x
 4 
∴
∴
∴
∴
∴
∴
∴
1
x4 = 64x
x4 – 64x = 0
x=0
or
x3 – 64 = 0
x(x3 – 43) = 0
x=0
or
x=4
Substituting for x in (i), we get y = 0 or y = 4
O ≡ (0, 0) and A ≡ (4, 4)
Draw AB perpendicular to X-axis
B ≡ (4, 0)
Volume of solid generated by revolving the
area between x2 = 4y and y2 = 4x about Xaxis = [Volume of solid generated when area
between y2 = 4x, x = 0 and x = 4 is revolved
about X-axis] – [Volume of solid generated
when area bounded by x2 = 4y is revolved
about X-axis from x = 0 and x = 4]
4
4
∫
4
4
∫
∫
0
0
= π 4x dx – π
4
1 1

2π
= π  – – 0 – 0 =
cubic units
3
5
15


Q-5) Find the volume of the solid obtained by
revolving about the Y-axis, the region
bounded by the curve x2 + y2 = 4 and the
line x = y 3 in the first quadrant.
Ans.
Y
2
2
 x2 
= π 4x dx – π 
 dx
4 
0
0
∫
 x3 x5 
–
= π

5 0
 3
x4
dx
16
 x5 
 4x 2 
= π

 – π
16 ( 5 ) 
 2 0
3,1
1
X′
–2 –1
1
–1
2
X
–2
3, 1
4
Y′
0
Application of Definite Integration
Mahesh Tutorials Science
14
Given equation of curve is x2 + y2 = 4
∴
2
2
x =4–y
Required volume
... (i)
b
=
And equation of line is x = y 3
∴
2
2
x = 3y
∴
y2 = 1
∴
=
=
Whe n y = 1, x = 3 and when y = –1,
x=– 3
∴
Thus the point of intersection is
(
)
3,1 and
)
3, –1 .
0
3
∫
∫x
0
0
x3
π 36x –
3
2
dx
3
0


33
π 36 ( 3 ) –
– ( 0 – 0 )
3


= π [36 (3) – 32]
Required volume = 99π
π cubic untis
to x =
2
2
Q-2) Find the volume of solid generated by
revolving by the curve y = sin x, from x = 0
Required volume
= π
a
= π 36 dx –
y = ±1
(
∫ (36 – x ) dx
3
4 – y2 = 3y2
4 = 4y2
∫
... (ii)
Now, to determine the point of intersection,
solving equations (i) and (ii), we get
∴
3
π y 2 dx = π
π
about the X-axis.
2
1
∫ ( 4 – y ) dy + π ∫ 3y dy
2
2
1
Y
0
3
1
 y3 
3

= π 4y – y  + π 3  
3 1

 3 0

8 
1 
= π  8 –  –  4 –   + π [1 – 0 ]
3 
3 

Ans.
X
0
x =
8 1

= π 8 – 4 – +  + π
3 3

7

= π 4 –  + π
3

5 
= π +π
3
Equation of curve is y = sin x, x = 0 to x =
π/ 2
π
8π
Required volume =
cubic units.
3
GROUP B – HOMEWORK PROBLEMS :
Q-1) Find the volume of solid generated by
rotating the area bounded by x2 + y2 = 36
and the lines x = 0, x = 3 about X-axis.
Ans. Equation of curve is x2 + y2 = 36
i.e. y2 = 36 – x2 and x = 0, x = 3
Application of Definite Integration
π
2
Required volume
=
∴
π
2
∫y
2
dx
0
π/ 2
=
π
∫
π/ 2
sin2 x dx = π
0
 1 – cos 2x
2
∫ 
0

 dx

π/2

π  sin2x 
π π  sinπ 
x–
=  –
–0–0
=



2
2 0 2 2  2 

π /2
=
π
sin 2x 
x–

2
2  0
=
π2
cubic units
4
=

π  π  sin π 
– 0 – 0
 –

2 2  2 

Mahesh Tutorials Science
15
Q-3) Find the volume of sphere generated by
rotating semicircle represented by
2
V = 2π
1
∫ 4 (4 – x )
2
dx
0
x2 + y2 + 2x + 4y = 4
2
Y
Ans. X ′
X
A
(3, 0)
0
A′
(–3, 0)
Given semicircle is x2 + y2 + 2x + 4y = 4
Expressing it in centre-radius form,
x2 + 2x + 1 + y2 + 4y + 4 = 4 + 1 + 4
(x + 1)2 + (y + 1)2 = 9
Its centre is (–1, –2) and radius is 3.
We shift the origin at (–1, –2).
∴
∴
∴
π
8

4 ( 2) – – 0 – 0 
2 
3

=
8π
cubic units
3
ordinates x = 0 and x = π . If this area is
4
revolved about X-axis, find volume of solid
so generated.
π /4
3
Y =9–X
The semicricle area is revolved about X-axis.
Required volume
0
π /4
=
∫ tan x dx = ( log sec x )
∫
π /4
0
0
= log sec
1
=
∫ y dx, where y = tan x
Ans. Required area =
i.e. Put x + 1 = X, y + 2 = Y
X2 + Y2 = 9
2
=
Q-5) Find the area bounded by y = tan x and
Y′
∴
=
2π 
x3 
 4x –

4 
3 0
π
– log sec 0
4
π Y 2 dX
= log 2 – log1
–3
1
= 2π
∫ (9 – X )
2
dX
= log 2 =
1
log 2 sq units.
2
... [ ∵ log 1 = 0]
0
π /4
 1
= 2π 9 1dX –
 0

∫

X 2 dX 

0

3
∫
Reqquired volume = π
∫y
2
dx ,
0
where y = tan x
3

 X3  
3

= 2π 9 ( X )0 – 
 
3 0 




π /4
=
π
∫ tan
2
x dx
0

 27  
= 2π 9 ( 3 ) –  3  



= 2π
π [27 – 9]
= 36π
π cubic units
π /4
=
π
∫ (sec
2
)
x – 1 dx
0
π /4
=
π [ tan x – x ]0
=


π π
π  tan –  – 0 
4 4


=
π

π 1 – 
4

=
π
( 4 – π ) cu units.
4
Q-4) Find the volume of the solid obtained by
revolving the area of the ellipse
x2 + 4y2 = 4 about the X-axis.
Ans. Equation of curve is x2 + 4y2 = 4
∴
1
y2 =
(4 – x2)
4
Required volume,
Application of Definite Integration
Mahesh Tutorials Science
16
BASIC ASSIGNMENTS (BA) :
Q-2) Find the area of region bounded by
parabola y2 = 4ax from to x = 0, x = b and
X-axis
BA – 1
Ans.
x2 y2
+
=1
Q-1) Find the area of ellipse
25 4
Y
y2 = 4ax
Ans. By the symmetry of the ellipse, its area is
A
equal to 4 times the area of the region OABO.
Clearly for this region, the limits of integration
are 0 and 5.
C
O
Y
X
B
O
X′
A (5, 0)
B
x=b
Y′
X
y2 = 4ax
∴ y =2 a
Y′
Required area = area of the region OBCAO
= 2 (area of the region OCAO)
From the equation of the ellipse,
b
y2
x 2 25 – x 2
=1 –
=
4
25
25
∴ y2 =
4
25 – x 2
25
(
∫
= 2 y dx
0
)
b
b
∫
=2 2 a
x dx = 4 a
0
In the first quadrant, y > 0
∴y=
x
∫
x dx
0
b
 x 3/2 
8 a 3/2
b
=4 a 
– 0 
 =
3 
 3 / 2 0
2
25 – x 2
5
∴ area of ellipse = 4 (area of region OABO)
5
=
5
2
= 4 y dx = 4
25 – x 2 dx
5
∫
∫
0
0
8
b ab sq units.
3
Q-3) Find area of region lying between y2 = x
and x2 = y
5
8 x
25
 x 
= 
25 – x 2 +
sin –1   
5 2
2
 5 0
Ans.
2
x = 4y
Y
 5

25

25 – 25 +
sin–1 (1)  – 

2
8  2


=

5
25
0

25 – 0 +
sin–1 0 


2
2


A
B
X′
O
D
(4, 0)
8 25 π
= .
. = 10 π sq units.
5 2 2
Y′
Application of Definite Integration
y 2 = 4x
C
X
Mahesh Tutorials Science
17
For finding the points of intersection of the
To find the points of intersection of line
two
parabolas, we equate the values of y2 from
2y = 3x + 12 and the parabola 4y = 3x2.
their equations.
From the equation x2 = y, y2 = x4
equations,
Equating the values of y from the two
we get,
∴ x4 = x x4 – x = 0
∴ x(x3 – 1) = 0
3x 2 3x +12
=
4
2
∴ x = 0, x = 1
When x = 0, y = 0
∴ 6x2 = 12x + 48
When x = 1, y = 1
∴ the points of intersection are O(0, 0) and
A(1,1).
Required area = area of the region OBACO
= (area of the region ODACO)
– (area of the region ODABO)
∴ x2 – 2x – 8 = 0
∴ (x – 4)(x + 2) = 0
∴ x = 4, x = –2
3 ( 4 ) +12
When x = 4, y =
2
= 12
Now, area of the region ODACO
= area under the parabola y2 = x, i.e., y = x
1
=
∫
0
When x = –2, y =
3 ( –2 ) +12
2
=3
1
 32 
x 
x dx = 
3 /2

0
∴ the points of intersection are D (–2, 3) and
B(4, 12).
Required area = area of the region OABCDO
2
2
= (1 – 0 ) =
3
3
= (area of the region EFBCDE)
– (area of the region EFBAODE)
Area of the region ODABO
Now, area of the region EFBCDE
= area under the parabola x2 = y
1
=
= area under the line
1
 x3 
x 2 dx =  
 3 0
0
∫
4
∫ ydx , When 2y = 3x + 12,
=
1
1
= (1 – 0 ) =
3
3
–2
2 1 1
– =
∴ required area =
sq unit.
3 3 3
i.e., y =
4
Q-4) Find the area cut off from the parabola
∫
=
–2
3x +12
2
3x +12
3
dx =
2
2
4
∫ ( x + 4) dx
–2
4y = 3x2 by the line 2y = 3x + 12
4
Ans.
=

3  x2
+ 4x 

2 2
 –2
=
3
( 8 +16 ) – ( 2 – 8 )  = 45
2
Y
4y
B (4, 12)
x
=3
C
2
Area of region EFBAODE
= area under the parabola
(–2, 3) D
X′
2y
=
3x
+
12
E
(–2, 0)
F
(4, 0)
X
4
=
∫
y dx , Where 4y = 3x2, i.e., y =
–2
4
Y′
=
∫
–2
3x 2
3
dx =
4
4
3x 2
4
4
∫x
2
dx
–2
Application of Definite Integration
Mahesh Tutorials Science
18
4
=
3  x3 
1
  = ( 64 + 8 ) = 18
4  3  –2 4
a
a
∫
∫
0
0
A1 = y dx =
∴ required area = 45 – 18
(∵ y
= 27 sq units.
Q-5) Find the area of the region lying above the
2ax – x 2 dx
2
= 2ax – x 2 , y > 0
)
a
A1 =
∫
(
)
a 2 – x 2 – 2ax + a 2 dx
0
X-axis and included between the circle
x2 + y2 = 2ax and the parabola y2 = ax
a
=
Ans.
∫
2
a 2 – ( x – a ) dx
0
a
x – a
a2
x – a
2ax – x 2 +
sin–1
A1 = 

2
a 0
 2
a – a
a2
a – a
A1 = 
2a (a ) – a 2 +
sin–1

2
a 
 2
0 – a
a2
0 – a
2a ( 0 ) – 02 +
sin–1
A1 = 

2
a 
 2
A1 = 0 (a ) +
We shall first find the points of
intersection of the circle and parabola.
A1 = 0 +
Equation of the parabola: y2 = ax …(i)
Equation of the circle:-
Similarly,
x2 + y2 = 2ax ...(ii)
Substituting the value of y2 obtained
A2 =
from equation(i) in equation (ii), we get,
x2 + ax = 2ax
∴ x 2 – ax = 0
a2
a
a2
sin–1 0 + ( 0 ) –
sin–1 ( –1)
2
2
2
a2
a2
π
πa 2
( 0 ) + 0 –  –  =
2
2  2
4
a
∫
ax dx
(∵ y
2
= ax , y > 0
a
 32 
3
x 
2 1
2
A2 = a 
= a 2 . a 2 = a2
3


3
3
 2 0
∴ x(x – a) = 0
x = 0 and x = a
∴ From(iii), A = A1 – A2
When x = 0, y 2 = a ( 0 ) = 0
=
When x = a , y 2a (a ) = a 2 ⇒ y = ± a
Points of intersection of the circle and the
parabola are (0,0), (a, a) and (a, –a).
)
0
πa 2
2
2
π
– a2 =  – a2
4
3
4
3


2
π
∴ Required area =  –  a 2 sq. units.
3
4
∴ The region above x -axis and included
between the circle and theparabola is
BA – 2
OQPRO.
Let A = Required area; A1 = Area of region
Q-1) Find the area of the region enclosed by the
OSPRO; A2 = Area of region OSPQO
Then A = A1 – A2 ...(iii)
Now, the region OSPRO is bounded by the
arc of the circle, the x-axis and the ordi
nates x = 0, x = a.
curve y = sin x and y = cos x and Y-axis.
Ans. To find the point of intersection of the curves
y = sin x
and y = cos x. Equating the values of y
from both the equations, we get,
Application of Definite Integration
Mahesh Tutorials Science
∴ tan x = 1 = tan
sin x = cos x
∴ x =
19
π
4
dt
2
∴ y dy = –
When y = 0, t = 1
π
4
1
When y =
when x =
π
π
1
, y = sin =
4
4
2
∴ area of the region OAPDO
π 1 
∴ the point of intersection is P  ,
.
2
4
sin
y=
B
x
π 1 
P ,

4 2
D
=
Y′
1
1/2
π
1
+
4 2 2
∫
t
–
1
2
dt
1
1/2
=
=
π
1
+
–1
4 2
2
sx
co
π
4
X′ O
1  dt 
. –

t  2 
∫
.
 1 
1  t2 
+
4 2 2 1/ 2 

1
y=
A
1/2
π
–
4
2
1
=
Y
C
2
1
2
,t=
π/2 X
π
Area of the region DPBCD
= area under the curve y = cos x
Required area = area of the region OAPBCO
1
∫
=
= area of the region OAPDO
x dy where x = cos–1 y
1/ 2
+ area of the region DPBCD
Now, area of the region OAPDO
1
∫ cos
=
= area under the curve y = sin x
–1
y dy
1/ 2
1/ 2
=
∫
x dy, when x = sin–1 y
=
0
∫
1
1/ 2
∫
=
∫
sin–1y dy
1/
0
1/ 2
= sin–1 y 1dy 

 0
∫
1/ 2
∫
0
(
–
)
(
1/ 2
–
)

–
2
∫
 1 
sin–1 
–0–
2
 2
1– y
1/ 2
∫
0

–1
1 – y2
2
. y dy
1

 1 
1
cos –1 
 +
= 0 –
2
 2   1/

.y dy
y
1 – y2
y
∫
2
1 – y2
dy
Put 1 – y2 = t ∴ – 2y dy = dt
∴ y dy =
=
1
1

) ∫ 1dy  dy

) ∫ 1dy  dy
1
∫
0
1
(
1/ 2
1/ 2
(
–
2
 d
cos –1 y

dy

2
–1
=  cos y .y 1/
 d
sin–1 y

dy

–1
=  sin y y  0
1
cos –1 y 1dy 

1/
dy
When y =
–dt
2
1
2
,t=
1
2
2
Put 1 – y = t ∴ – 2y dy = dt
Application of Definite Integration
Mahesh Tutorials Science
20
When y = 1, t = 0
5
=
∴ area of the region DPBCD
=
∫
.
0
–π
=
 1 
1  t2 
–
4 2 2 1/ 2 

1/2
0
0
25 – x 2 dx
5
=
∫
x
25
 x 
4
25 – x 2 +
sin–1   
2
2
 5  0

 5

25

25 – 25 +
sin –1 (1) 


2
2



4
=

25
0

–1
–
25 – 0 +
sin ( 0 ) 

0
2


0
–π
= 4.

1 
–π
1
– 0 –
+
=
4 2 
2 4 2
2
25 π
. = 25 π sq units.
2 2
–π
∴ required area =
=
∫
1
=
–
1
–
t 2 dt
4 2 2 1/2
=
∫
0
π
1  dt 
+
– 2 
4
2
t


1/2
1
–
5
4 y dx = 4
2
2
–1=
(
π
4 2
+
1
2
–1–
Q-3) Find the area of the region bounded by the
π
4 2
+
parabola y = x2 and the line y = x in the
first quadrant.
1
2
Ans.
)
2 – 1 sq unit.
2
y=x
Y
B (1,1)
Q-2) Find the area of circle x2 + y2 = 25
C
Ans.
A
O
D (1,0)
X
=
x
X′
y
Y
Y′
B
The vertex of the parabola y = x2 is at the
origin O (0, 0).
X′
O
A (5, 0) X
To find the points of intersection of line and
the parabola.
Equating the values of y from the two
equations,
we get,
Y′
x2 = x
∴ x2 – x = 0
∴ x (x – 1) = 0
By the symmetry of the circle, its area is equal
to 4 times the area of the region OABO.
Clearly for this region the limits of integration
are 0 and 5. From the equation of the circle,
y2 = 25 – x2. In the first quadrant y > 0
2
∴ y = 25 – x
∴ area of the circle = 4 (area of region OABO)
∴ x = 0, x = 1
When x = 0, y = 0
When x = 1, y = 1
∴ the points of intersection are O(0, 0) and
B (1, 1).
Required area = area of the region OABCO
= (area of the region ODBCO) –
(area of theregion ODBAO)
Application of Definite Integration
Mahesh Tutorials Science
21
Now, area of the region ODBCO
Required area = area of the region ABCDA
= area under the line
= (area of the region AOECDA)
– (area of the region AOECBA)
1
=
∫
y dx , where y = x
Now, area of the region AOECDA
0
= area under the line
1
=
1
x 
x dx =  
 2 0
0
2
∫
2
=
∫ y dx where y = 2x + 1
0
=
1
1
(1 – 0 ) =
2
2
2
=
Area of the region ODBAO
∫
= (4 + 2) – 0 = 6
= area under the parabola
Area of the region AOECBA
1
=
2
 x2

( 2x +1) dx = 2. + x 
2

0
0
∫ y dx, where y = x
2
= area under the parabola
0
2
1
=
∫
0
1
x3 
x dx =  
 3 0
2
=
1
1
(1 – 0 ) =
3
3
=
∫ y dx where y = x
2
+1
0
2
=
1 1 1
– = sq unit.
∴ required area =
2 3 6
0
)
14
8

=  + 2 – 0 =
3
3

Q-4) Find the area of the region bounded by the
2
parabola y = x + 1 and line y = 2x + 1
Ans. The equation of the parabola can be written
as x2 = y – 1. Hence its vertex is at A(0, 1)
and it opens upwards. To find the points of
intersection of line and the parabola.
∫(
2
 x3

x 2 +1 dx = 
+ x
 3
0
∴ required area = 6 –
14 4
sq units.
=
3
3
Q-5) Find the area of the region enclosed
between the circles x2 + y2 = 1 and
Equating the values of y from the two
(x – 1)2 + y2 = 1
equations, we get,
Ans.
x 2 + 1 = 2x + 1
∴ x2 – 2x = 0 ∴ x (x – 2) = 0
Y
∴ x = 0, x = 2
When x = 0, y = 0 + 1= 1
When x = 2, y = 2 × 2 + 1= 5
G
∴ the points of intersection are A(0, 1) and
C(2, 5).
1
X
C
A
B
=
2x
+
P
O
y
2
+1
y=x
X′
E
D
X′
Y
F
C (2,5)
D
B
(0,1)
A
O
E (2,0)
Y′
X
To find the points of intersection of the
circles x2 + y2 = 1 and (x – 1)2 + y2 = 1
Y′
Application of Definite Integration
Mahesh Tutorials Science
22
Equating the values of y2 from the two
equations,
We get,
1 – x2 = 1 – (x – 1)2
2
∴ (x – 1) = x
2
∴ –2x + 1 = 0
2
∴ x – 2x + 1 = x
∴x=
2
1
2
1/2
 ( x – 1)
1
2
 x – 1 
1 – ( x – 1) + sin–1 

= 2
2
 1 0

 1

2
– 2
1
1 
 1
–1 
1 –  –  + sin  –  

=  2
2
 2
 2 


1
 1

– –
1 – 1 + sin−1 ( –1) 
2
 2

2
1
1 3
1
When x = , y 2 = 1 –   = 1 – =
2
4 4
2
= –
3
∴y = ±
2
1
3
∴ the points of intersection are  ,
 2 2 


1 – 3 
and B  ,
.
 2 2 


1 3 1 π
1 π
.
+ – +0 – – 
4 2
2 6
2 2
= –
3
π π π
3
–
+ = –
8 12 4 6
8
Area of the region PDEFP
= area under the circle x2 + y2 = 1 from x =
Re qu ire d are a = are a of th e re gi on
to x = 1
1
OABCDEFGO
=
= 2 (area of the region ODEFGO)
∫ y dx, where y
2
2
= 1 – x2, i.e., y = 1 – x
1/2
= 2 (area of the region OPFGO
1
+ area of the region PDEFP)
=
Now, area of the region OPFGO
2
∫
1 – x 2 dx
1/2
2
= area under the circle (x – 1) + y = 1 from
x = 0 to x =
1
1
x

1 – x 2 + sin –1 x 
= 
2
2
1/2
1
2
1/2
=
∫ y dx , where y
2
= 1 – (x – 1)2

1
1 1

 1
–1
–1  1 
= 0 + 2 sin (1)  –  4 1 – 4 + 2 sin  2  

 
  
0
= 1 – x2 + 2x – 1
=
1 π 1 3 1 π
. – .
– .
2 2 4 2
2 6
=
π
3
π
π
3
–
–
= –
4
8 12 6
8
= 2x – x2
∴ y = 2x – x 2
1/2
∫
=
π
3 π
3
+ –

∴ required area = 2  –
8
6
8 
6
2x – x 2 dx
0
1/2
∫
=
(
)
1 – x 2 – 2x +1 dx
0
1/2
=
∫
1
2
2
1 – ( x – 1) dx
0
Application of Definite Integration
=
π
3 π
3
–
+ –
3
4
3
4


=  2π – 3  sq units.
 3

2 

Mahesh Tutorials Science
23
∴ volume of the solid obtained by revolving
the ellipse about the major axis, i.e. X-axis
BA – 3
Q-1) Find the volume of solid generated by
revolving the area bounded by curve
y = sin x, from x = 0 to x =
a
∫
= π y 2 dx
–a
π
abount X2
a
= π
axis.
b2 2
a – x 2 dx
a2
∫ (
–a
)
π /2
Ans. Required volume= π
∫ y dx, where y = sin x
2
0
π /2
= π
∫
π /2
sin2 x dx = π
0
= π
2
∫
0
 1 – cos 2x 

 dx
2


π /2
 π2
π  π

cu units.
 – 0  – 0  =
2  2

 4
2
)
– x 2 dx
0
a
2πb 2
a2
 2
x3 
a x –

3 0

=
2πb 2
a2
 3 a 3  2πb 2 2a 3
.
a –
=
3 
3
a2

=
4
πab 2 cu units.
3
Q-3) The area bounded by the curve y2 = x2
Q-2) Find the volume of solid generated by
revolving the area of ellipse
∫ (a
=
0
π /2
a
...[ ∵ f (x) = a2 – x2 is an even function]
∫ (1 – cos 2x ) dx
π
sin 2x 
= x –
2
2  0
=
2πb 2
=
a2
x2 y2
+
=1
a2 b2
(1 – x2) between x = 0 and x = 1 is rotated
about X-axis. Obtain the volume of solid
so generated.
1
∫
about major axis.
2
Ans. Required volume = π y dx ,
Ans. From the equation of the ellipse,
0
where y2 = x2 (1 – x2)
y2
x2 a2 – x2
=
1
–
=
b2
a2
a2
1
∫ (
)
= π x 2 1 – x 2 dx
b2
∴ y = 2 a2 – x2
a
2
(
)
0
1
= π
Y
∫ (x
2
)
– x 4 dx
0
1
 x3 x5 
–
= π

5 0
 3
B (0, b)
X ′ A′ (– a, 0)
O
A (a, 0) X
B′ (0, –b)
1 1

= π  – – 0
3
5


=
2π
cu units.
15
Y′
Application of Definite Integration
Mahesh Tutorials Science
24
Q-4) Find the volume of the solid obtained by
Y
the revolving about X-axis and the region
bounded by the y2 = 4x and lines x = 4 and
x=5
y=
α
5
∫
2
Ans. Required volume = π y dx , where y2 = 4x
O
4
5
A
x
h)
(r/
r
A
(h,0)
h
5
∫
∫
4
4
= π 4x dx = 4π x dx
B
Y′
5
5
 2
= 4π  x  = 2π  x 2 


4
 2 4
The the slope of the line OA
= m = tan α = r/h
= 2π (25 – 16)
∴ equation of the line OA is y = mx
i.e. y = (r/h) x
= 18π cu units.
Q-5) Find the volume of the solid obtained by
the revolving about X-axis and the region
bounded by the curve, xy = 1 and lines
x = 1 and x = 2
Revolving this line about the X-axis, we get
the required cone.
h
∫
2
∴ volume of the cone = π y dx
0
Ans. From the equation of the curve,
h
1
1
2
y = , i.e. y = 2
x
x
= π
2
2
2
∫
∫
1
1
2
∴ required volume = π y dx = π
1
dx
x2
2
 x -1 
–2
= π x dx = π 

 –1 1
1
∫
2
1 
1

= – π   = – π  – 1
 x 1
2

=
X
π
cu units.
2
BA – 4
=
h
r2 2
πr 2  x 3 
. x dx = 2  
2
h
h  3 0
0
∫
πr 2 h 3 1 2
.
= πr h cu units.
3
h2 3
Q-2) Find the volume of solid generated by
revolving the area bounded by curve
9x2 – 4y2 = 36, from x = 2 to x = 4 about Xaxis
Ans. From the equation of the curve
(i.e., hyperbola),
y2 =
∴
9 2
x –4
4
required volume
(
)
4
Q-1) Find the volume of cone of height ‘h’ and
base radius ‘r’
Ans. Let OAB be a cone of height h, base radius r
and semi vertical angle α. Take O the vertex
of the cone as the origin, the X-axis along
the axis of the cone and the line OY through
O and perpendicular to the X-axis as the Yaxis.
Application of Definite Integration
∫
2
= π y dx = π
2
4
9
∫ 4 (x
2
)
– 4 dx
2
4

9π  x 3
– 4x 

=
4  3
2
=
  23

9π  43
– 16  – 
– 8 

4  3
  3
 
=
9π  64
8

– 16 – + 8 

4  3
3

=
9π 32
×
= 24π cu units.
4
3
Mahesh Tutorials Science
25
Q-3) Find the volume of the solid obtained by
π

= π 1 – 
4

revolving about the X-axis, the region
bounded by x2 + y2 = 25 in interval from
x = 0 and x = 3
Ans. From the equation of the curve (i.e. circle),
y2 = 25 – x2
the revolving about the X-axis and the
region bounded by the curve
∫
2
∴ required volume = π y dx
0
9y2 = x2 (a – x)
Ans. Put y = 0
∴ x = 0 or x = a
we have,
3
∫ (25 – x ) dx
2
0
3

x3 
x
π
25
–

= 
3 0

b
∫
V = π y 2 dx
a
= π [(75 – 9) – 0]
= 66π cu units.
a
= π
Q-4) Find the area bounded by y = tan x and
ordinates x = 0 and x =
π
π /4
∫ y dx , where y = tan x
0
π /4
∫
0
π /4
tan x dx = ( log sec x ) 
0
∫
0
3
4
= π  ax – x 
9 3
4 0
3

π  a (a )
a4
= 
–
– 0 + 0
9 3
4


=
π
– log sec 0
4
π a 4 a4 
–


9 3
4 
πa 4
=
9 × 12
= log 2 – log1
= log 2 =
dx
π
ax 2 – x 3 dx
=
9
0
= log sec
9
a
so generated.
Ans. Required area =
∫
x 2 (a – x )
a
. If this area is
4
revolved about X-axis, find volume of solid
=
π
( 4 – π ) cu units.
4
Q-5) Find the volume of the solid obtained by
3
= π
=
1
log 2 sq units.
2
... [ ∵ log 1 = 0]
=
πa 4
cubic units.
108
π /4
Reqquired volume = π
∫y
2
dx ,
0
where y = tan x
π /4
= π
∫ tan
2
x dx
0
π /4
= π
∫ (sec
2
)
x – 1 dx
0
π /4
= π [ tan x – x ]0


π π
= π  tan 4 – 4  – 0 



Application of Definite Integration