Example 2 - EngineeringDuniya.com
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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
ax
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
2a 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
ax
dx
3a 2x
x
dx
3
2a 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
8a
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 2y 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
2y
Re quried surface area =
2yds
0
2
ds
2y d
d
0
2
2
a(1 cos )2a sin d
2
0
2
64a
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 2y d
d
0
0
2r 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.