Transcript Example 2 - EngineeringDuniya.com

APPLICATION OF
INTEGRATION

The integration can be used to determine
the area bounded by the plane curves, arc
lengths volume and surface area of a
region bounded by revolving a curve
about a line.
I. AREA OF THE PLANE REGION

We know that the area bounded by a
Cartesian curve y = f(x), x – axis,
between lines x = a & x = b given by
b
Area   f (x)dx
a
Y
b
Area   f (x)dx
a
(0, 0)
x
 = c2
Y
 = c1
r=f()
X
0
r
o

Area=(1/2)r2 



Example 1:
To find the area lying between the parabola
y =4x – x2 and the line y = x.
The required area = ( The area bounded by
the parabola y =4x – x2, the x – axis, in
between lines x = 0 and x = 3) – the area
bounded by y = x, x – axis, in between lines
x = 0 and x = 3.
3
Area 

x 0


3
9
4x  x dx   xdx 
2
0
2
Y
(3,3)
(0,0)
x=3
X

Example 2 :
To Sketch and find the area bounded by
the loop of the curve 3ay2 = x( x – a)2.
The curve is symmetric about the x – axis,

x x  a
x(x  a)
y 
y
3a
3a
Therefore y is defined if x 0.


2
2




The curve intersects x – axis at (0,0) &
(a, 0)
Therefore loop in formed between
these point.
x = 0 i.e., y – axis is tangent to the
curve at the origin.
The curve does not have any
asymptotes.
Y
(a,o)
0
Graph of the given curve
X


The area bounded by the loop of the
curve
= 2 area bounded by the portion of the
loop of the curve, x – axis lying in the first
quadrant
a
a
0
0
 2  ydx  2  
8

a2
15 3
x (x  a)
dx
3a




Example 3:
To find the area inside the cardioide
r = a ( 1+ cos) and the circle r = 2 cos
Required area is given by
=  / 2
X
X
X

Area
1 2
a 2
2
 2   a 1  cos   d 
 2
2
0







  a 2 1  cos 2   2 cos  d  a 2
0

 2


2
 a 2  2  1d  2  cos 2 d   cos d  a 2 


0
0
0



1 

2
 a  2  2 .  0 2
2 2
 2

2
   a
a  
2
2
2






Example 4:
To find the area bounded by the curve
y2(a-x) = x3 and its asymptote.
X = a is the asymptote to the curve.
The required area is given by
Area = 2 The area bounded by the
curve and the asymptote lying in the
first quadrant.
Y
a
 2  ydx
X
0
a
o
x
dx
ax
 2 x
0
X=a
Put x = a sin 2 

 4 a2
2

0
2
3
1

3

a
sin 4   4 a 2 . .

4 22
4
EXERCISES



1. Find the area bounded by the one
arch of the cycloid x = a ( - sin ) ,
y = a ( 1- cos ) and its base
2. Find the area of the region lying
above x – axis, included between the
circle x2 + y2 = 2ax and the parabola
y2 = ax.
3. Find the area between the curve
x ( x2 + y2) = a(x2 – y2) and its
asymptote.

4.Find the area bounded by the curve .
2
x



3
y
2
3
a
2
3
5.Find the area bounded by the loops of the
curve r2 = a sin 
6. Find the area inside r = a ( 1 – cos )
and outside r = a sin 
7: Find the area common to the curves
r = a ( 1+ cos) and r = a ( 1 - cos)



8. Find the area inside r = a and
outside r = 2acos.
9. Find the area bounded by the loops
of the curve x3 + y3 = 3axy.
10. Find the area bounded by the
loops of the curve r = a sin3.
RECTIFICATION-LENGTH OF THE
PLANE CURVE

The rectification is the process of
determining the length of the arch of a
plane curve. We know that the
derivative of the arc of length of a
curve is given by
2
ds
 dy 
ds 
dx  1    dx for a cartesian curve y = f(x).
dx
 dx 
2
ds
2  dr 
ds  d  r    d for a polar curve r = f()
d
 d 
2
2
ds
 dx   dy 
ds  d      d for a parametric curve
d
 d   d 
x = f()and y=g()


The length of the arc of the curve is given by
Arc length s =

b
a
ds
2
b
 dy 
1    dx for a cartesian curve y = f(x)
 dx 

a
b
=
b
b
ds
 d d  
a
a
b
2
 dx   dy 
     for a parametric
 d   d 
curve x = f() & y  g()
2
ds
 dr 
d=  r 2    d for a polar curve r = f()
d
 d 
a
a



To find the length of the arc of the
parabola x2 = 4ay measured from
the vertex to one extremity of the
latus rectum.
Here
x2
y

4a
dy x

2a
dx
2
ds
1
 dy 
Therefore
 1 
 
dx
2a
 dx 
4a 2  x 2

The required arc length
2a


0
ds
dx 
dx
2a

0
1
4a 2  X 2 dx
2a
2a
2


1 x
4a
2
2
 
4a  x 
sinh 1 x  
a 
2a  2
2
0
 
a


2  log 1  2


To find the perimeter of the curve
2


x
3
y
2
3
a
2
3
The parameter equation of this curve is
x = a cos3, y = a sin3.
2
ds  dx   dy 
    
d  d   d 
 3a cos  sin 
2

Therefore Perimeter of the curve

2
π
2
ds
4 
 4  3a sinθ cosθ dθ
d

0
0
=6a


To find the perimeter of the curve
r = a ( 1+ cos).
dr
 a sin 
d
ds

d
r
2
 2a cos 
 dr 


 d 
2
2

ds
Therefore Perimeter of the cardioide =2  d
d

0

 2 2a cos  d
2
0
 8a
EXERCISES




1.Find the length of the arc of one arch of
the cycloid x = a(+ sin), y = a ( 1- cos ).
2.Find the length of the loop of the curve
3ay2 = x (x – a)2.
3.Find the length of the arc of the curve of
the centenary y = c cosh(x/c) measured
from the vertex to any point (x ,y).
4.Find the length of arc of the loop of the
curve r r2 = a2 cos2.
VOLUME OF REVOLUTION
Let a curve y = f(x) revolve about x–axis.
Then the volume of the solid bounded by
revolving the curve y = f(x), in between
the lines x = a and x = b, about x – axis is
given by
b
volume   y dx
2
a

If the curve revolves about y – axis, the
volume is given by
b
volume   x 2dy
a


Examples :
1.To find the volume of the solid obtained
by revolving one arch of the curve x = a
(+sin) , y = a (1 + cos) about its base

X- axis is the base of the curve
Y
-a

o
0
Therefore the required volume
a
X
a


a
 a

y 2 dx  2   y 2dx
0
dx
 2  y
d
d

0
2

 2  a 2 (1  cos ) 2 a(1  cos )d
0

 2a 3  (1  cos )3d
0
2 3
 5 a



2. To find the volume bounded by
revolving the curve y2(a – x ) = x3 about
its asymptote.
X = a is the asymptote to the curve.
Shifting the origin to the point (a, 0), we
get the new coordinates X = x – a &
Y = y – 0 = y.
Then the volume bounded by revolving
the curve about the asymptote is given
by

volume  2

x 2dy
y 0

2 
 x  a
2
dy
y 0
a
2 
 x  a
2
 x  a
2
x 0
a
2 
x 0
 x x
d
 ax

dx

3a  2x 

x
dx
3
2a  x  2
Put x  a sin , x  0    0
x=a  = 
2
2
Volume

 2

2
    6a 3 cos 2  sin 2 d   4a 3 cos 2  sin 4 d 


0
 0

1 3 1 
3 1 1 
 a  6. . .  4. . . . 
6 4 2 2
 4 2 2
2 a 3

4


2.To find the volume of the solid bounded
by revolving the cardioide r = a(1+cos )
about the initial line
Required volume
2a


y 2dx
x 0
where x = r cos = a(1-cos )cos 
y=r sin  = a (1+ cos )sin .
x = 2a  = 0, x = 0  =
Therefore the volume
0
=
 a 2 (1  cos ) 2 sin 2  d  a(1  cos )cos  



 a  sin (1  cos ) (1  2cos )d
3
8a

5
3
0
3
2
PROBLEMS


The loop of the curve 3ay2=x(x – a)2
moves about the x – axis ; find the
volume of the solid so generated.
Find the volume of the spindle shaped
solid generated by revolving the curve
about x – axis .
SURFACE AREA OF REVOLUTION
The area of the surface of the solid obtained
by revolving about x – axis, the arc of the
curve
y = f(x) intercepted between the points
whose abscissa are a and b , is given by
b
ds
surface area   2y dx
dx
a



Examples :
1.To find the area of the surface of the
solid generated by revolving on arch of
the curve x= a (  – sin), y = a(1 –
cos) about its base.
X =a( -sin ) ,y = a(1 - cos )

2
 dx 
 dy 

 

d

d





 2a sin 
2
ds

d
2
2y
Re quried surface area =

2yds
0
2
ds
  2y d
d

0
2
 2

a(1  cos )2a sin  d
2
0
2
64a

3
2.To find the surface of the solid formed by
revolving the curve r = a(1+cos) about
the initial line ,
 dr 
r 

 d 
 2a cos 
2
ds

d
2
2

ds
Therefore surface area   2y d
d

0



0
2r sin  2a cos  d
2
32 2
 a
5
EXERCISES




1.Find the area of the surface of the solid
generated by revolving the arc of the parabola
y2 = 4ax bounded by its latus rectum about
x –axis.
2.Find the area of the surface of revolution
about the x – axis the curve x2/3 +y2/3 = a2/3 .
3.Find the total area of surface of revolution of
the curve r2 = a2cos2 about the initial line.
4.Find the area of the surface of revolution of
the loop of the curve 3ay2=x(x-a)2 about the x –
axis.