Mathematics Exam > Mathematics Notes > Topic-wise Tests & Solved Examples for Mathematics > Line Integral
Line Integral | Topic-wise Tests & Solved Examples for Mathematics PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 2
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( t
i t
) ( t it ) dt
? ( t
i t
i t
t
) dt
i ? t
dt
i 4
t
5
i
i
3
Ex.1\Pg.46 Evaluate , where f (z) =x
2
+iy
2
and C is given by
z ( t) t i t t
Sol
n
:- z ( t) t it z ( t ) i
But z = x+iy
x =t , y=t
? f ( z ) ( x(t ) )
2
+i (y(t))
2
=t
2
+ i t
2
? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( t
i t
) ( i ) dt
i ? t
dt
i 4
t
3
5
3
i
NOTE: Curve in Que 2 & 3 both are smoothly equivalent by thm 4.5
Qu e 3 ? f
w her e f ( z ) z
C is g iven by z ( t ) sin t i cos t , t
S oltio n f ( z ) z
x iy x i y
x
y
x
x
y
y
x
y
z(t) = sin t + i cost z ( t) cos t – i sin t
but z = x +i y
? x sin t , y = cos t
? x
2
+ y
2
=1
f(z(t)) = sin t –i cos t
? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( sin t i cost ) ( cost i sin t ) dt
? ( sin t cost i ) dt
Ex pg E v aluate ? f
w her e f ( z ) z
C is g iven by z ( t ) R cos t i R sin t , t , R
S olu ti on f ( z ) z
x i y
x i y x
y
z ( t ) R cos t i R sin t z
( t ) R sin t i R cos t
But, z =x+iy
? ( ( ) ) R cos R
R sin R
cos R
sin R
? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
?
cos i sin R
( R sin t i R cos t ) dt
? ( C ost iS in t ) ( S in t iCost ) dt
? ( sin t cos t i sin
t i cos
t cos t sin t ) dt
? ( sin
t cos
t ) dt
i ? . dt
i ( t )
i
T he cu r v e i s sai d to b e smooth iff z ( t) , excep t a t a f in it e n o. of p oi nt s.
Page 3
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( t
i t
) ( t it ) dt
? ( t
i t
i t
t
) dt
i ? t
dt
i 4
t
5
i
i
3
Ex.1\Pg.46 Evaluate , where f (z) =x
2
+iy
2
and C is given by
z ( t) t i t t
Sol
n
:- z ( t) t it z ( t ) i
But z = x+iy
x =t , y=t
? f ( z ) ( x(t ) )
2
+i (y(t))
2
=t
2
+ i t
2
? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( t
i t
) ( i ) dt
i ? t
dt
i 4
t
3
5
3
i
NOTE: Curve in Que 2 & 3 both are smoothly equivalent by thm 4.5
Qu e 3 ? f
w her e f ( z ) z
C is g iven by z ( t ) sin t i cos t , t
S oltio n f ( z ) z
x iy x i y
x
y
x
x
y
y
x
y
z(t) = sin t + i cost z ( t) cos t – i sin t
but z = x +i y
? x sin t , y = cos t
? x
2
+ y
2
=1
f(z(t)) = sin t –i cos t
? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( sin t i cost ) ( cost i sin t ) dt
? ( sin t cost i ) dt
Ex pg E v aluate ? f
w her e f ( z ) z
C is g iven by z ( t ) R cos t i R sin t , t , R
S olu ti on f ( z ) z
x i y
x i y x
y
z ( t ) R cos t i R sin t z
( t ) R sin t i R cos t
But, z =x+iy
? ( ( ) ) R cos R
R sin R
cos R
sin R
? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
?
cos i sin R
( R sin t i R cos t ) dt
? ( C ost iS in t ) ( S in t iCost ) dt
? ( sin t cos t i sin
t i cos
t cos t sin t ) dt
? ( sin
t cos
t ) dt
i ? . dt
i ( t )
i
T he cu r v e i s sai d to b e smooth iff z ( t) , excep t a t a f in it e n o. of p oi nt s.
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
i. e. iff x ( t) y ( t) ,
Smoothly equivalent Curve Curve C 1 g iven b y z z ( t) a t b is s . t . b . smo ot hly equivalent to Curve C 2 given
b y w w ( t) c t d . If there exist a on e t o one cont in uou sly d iff ma p p in g
? ,c , d - ,a , b - s. t .
(i) ? ( c) a , ? ( d ) b , ? ( t ) = ? t increasing
(ii) z ( ?( t) ) w ( t )
Que. 1 Prove that a smoothly equivalent curve is an equivalence relation.
Reflexive C given by z = z(t) is smoothly equivalent to itself.
S in ce ? can b e t aken a s i d en ti ty f un
c
.
Symmetric let C 1 be smoothly equivalent to C 2.
Let ? ,c, d - ,a , b - b e one to one sat isfy in g cod
n
?( c) a ?( d ) b & z ( ?( t ) ) w ( t) then
?
1
,a, b - ,c, d - w ill g iv es C 2 which is also smoothly equivalent to C 1
Transitive let C 1 : z=z(t) , a t b
C 2 w w ( t) c t d
C 3 r r ( t) e t f
If C 1 be smoothly equivalent to C 2
? ? a one to one map p i ng ? 1 ,c, d - ,a , b - s. t .
? 1 ( c) a, ? 1 ( d ) b z ( ? 1(t))=w(t)
and C 2 be smoothly equivalent to C 3 then
? 2 ( e ) c , ? 2 ( f ) d w ( ? 2(t))=r(t)
let ? ? 1 ° ? 2 then ? i s on e t o one such that
? ,e, f - ,a , b -
?( e) ( ? 1 ° ? 2 ) ( e) ? 1 (? 2 ( e) ) ? 1(c)=a
?( f ) ( ? 1 ° ? 2 ) ( f ) ? 1 (? 2 ( f ) ) ? 1(d)=b
also, z ( ?( t) ) z ( ( ? 1 ° ? 2)(t))
z ( ? 1 (? 2 ( t) ) ) w ( ? 2(t)) = r(t)
Therefore, C 1 is smoothly equivalent to C 3
Hence, a smoothly equivalent curve is an equivalence relation.
T hm . If C
C
are smo othly eq uiv ale nt cu r v e , then ? f
c
? f
c
Proof let C 1 :z = z(t) , a t b
C 2 :w = w(t) , c t d
Let C 1 & C 2 be smoothly equivalent, therefore there exists a mapping
? ,c, d - ,a , b - s. t .
?( c) a , ?( d ) b z ( ?( t ) ) w ( t)
Consider,
? ? f
? f ( z ) dz
? f ( z ( t ) ) z
( t ) dt
= , u
( z ( t) ) iv ( z ( t) ) -,x ( t ) iy ( t) -d t
= , u
( z ( t) ) x ( t ) d t v
( z ( t) ) y ( t) d t- i, u
( z ( t) ) y ( t ) d t v
( z ( t) ) x ( t ) d t …… …( i)
Consider,
? ? f
? f ( w ) dw
? f ( w ( t ) ) w
( t ) dt
= f
( z ( ? ( t) ) ) z (?( t ) ) ? ( t ) )
= , u
( z ( ? ( t) ) ) i v( z ( ?(t) ) ) - ,x ( ?( t ) ) i y ( ? ( t) ) -? ( t) d t
=[ u
( z ( ? ( t) ) x ( ? ( t) ) ? ( t) d t v
( z ( ? ( t) ) ) y ( ?( t ) ) ? ( t ) d t- i, u
( z ( ?( t) ) ) y ( ?(t) ) ? ( t) d t
+ v
( z ( ?(t) ) ) x ( ?( t ) ) ? ( t ) d t- ……( ii )
eq
n
(i) & (ii) are equal
by change of variable of real integral
ls o , ? x ( t )
dt ? y ( t )
???? y c ha n ge of va r ia b l e
Put ?(t) = r When
t =c then ?(c)=a & t =d then ?(d)=b
Page 4
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( t
i t
) ( t it ) dt
? ( t
i t
i t
t
) dt
i ? t
dt
i 4
t
5
i
i
3
Ex.1\Pg.46 Evaluate , where f (z) =x
2
+iy
2
and C is given by
z ( t) t i t t
Sol
n
:- z ( t) t it z ( t ) i
But z = x+iy
x =t , y=t
? f ( z ) ( x(t ) )
2
+i (y(t))
2
=t
2
+ i t
2
? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( t
i t
) ( i ) dt
i ? t
dt
i 4
t
3
5
3
i
NOTE: Curve in Que 2 & 3 both are smoothly equivalent by thm 4.5
Qu e 3 ? f
w her e f ( z ) z
C is g iven by z ( t ) sin t i cos t , t
S oltio n f ( z ) z
x iy x i y
x
y
x
x
y
y
x
y
z(t) = sin t + i cost z ( t) cos t – i sin t
but z = x +i y
? x sin t , y = cos t
? x
2
+ y
2
=1
f(z(t)) = sin t –i cos t
? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( sin t i cost ) ( cost i sin t ) dt
? ( sin t cost i ) dt
Ex pg E v aluate ? f
w her e f ( z ) z
C is g iven by z ( t ) R cos t i R sin t , t , R
S olu ti on f ( z ) z
x i y
x i y x
y
z ( t ) R cos t i R sin t z
( t ) R sin t i R cos t
But, z =x+iy
? ( ( ) ) R cos R
R sin R
cos R
sin R
? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
?
cos i sin R
( R sin t i R cos t ) dt
? ( C ost iS in t ) ( S in t iCost ) dt
? ( sin t cos t i sin
t i cos
t cos t sin t ) dt
? ( sin
t cos
t ) dt
i ? . dt
i ( t )
i
T he cu r v e i s sai d to b e smooth iff z ( t) , excep t a t a f in it e n o. of p oi nt s.
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
i. e. iff x ( t) y ( t) ,
Smoothly equivalent Curve Curve C 1 g iven b y z z ( t) a t b is s . t . b . smo ot hly equivalent to Curve C 2 given
b y w w ( t) c t d . If there exist a on e t o one cont in uou sly d iff ma p p in g
? ,c , d - ,a , b - s. t .
(i) ? ( c) a , ? ( d ) b , ? ( t ) = ? t increasing
(ii) z ( ?( t) ) w ( t )
Que. 1 Prove that a smoothly equivalent curve is an equivalence relation.
Reflexive C given by z = z(t) is smoothly equivalent to itself.
S in ce ? can b e t aken a s i d en ti ty f un
c
.
Symmetric let C 1 be smoothly equivalent to C 2.
Let ? ,c, d - ,a , b - b e one to one sat isfy in g cod
n
?( c) a ?( d ) b & z ( ?( t ) ) w ( t) then
?
1
,a, b - ,c, d - w ill g iv es C 2 which is also smoothly equivalent to C 1
Transitive let C 1 : z=z(t) , a t b
C 2 w w ( t) c t d
C 3 r r ( t) e t f
If C 1 be smoothly equivalent to C 2
? ? a one to one map p i ng ? 1 ,c, d - ,a , b - s. t .
? 1 ( c) a, ? 1 ( d ) b z ( ? 1(t))=w(t)
and C 2 be smoothly equivalent to C 3 then
? 2 ( e ) c , ? 2 ( f ) d w ( ? 2(t))=r(t)
let ? ? 1 ° ? 2 then ? i s on e t o one such that
? ,e, f - ,a , b -
?( e) ( ? 1 ° ? 2 ) ( e) ? 1 (? 2 ( e) ) ? 1(c)=a
?( f ) ( ? 1 ° ? 2 ) ( f ) ? 1 (? 2 ( f ) ) ? 1(d)=b
also, z ( ?( t) ) z ( ( ? 1 ° ? 2)(t))
z ( ? 1 (? 2 ( t) ) ) w ( ? 2(t)) = r(t)
Therefore, C 1 is smoothly equivalent to C 3
Hence, a smoothly equivalent curve is an equivalence relation.
T hm . If C
C
are smo othly eq uiv ale nt cu r v e , then ? f
c
? f
c
Proof let C 1 :z = z(t) , a t b
C 2 :w = w(t) , c t d
Let C 1 & C 2 be smoothly equivalent, therefore there exists a mapping
? ,c, d - ,a , b - s. t .
?( c) a , ?( d ) b z ( ?( t ) ) w ( t)
Consider,
? ? f
? f ( z ) dz
? f ( z ( t ) ) z
( t ) dt
= , u
( z ( t) ) iv ( z ( t) ) -,x ( t ) iy ( t) -d t
= , u
( z ( t) ) x ( t ) d t v
( z ( t) ) y ( t) d t- i, u
( z ( t) ) y ( t ) d t v
( z ( t) ) x ( t ) d t …… …( i)
Consider,
? ? f
? f ( w ) dw
? f ( w ( t ) ) w
( t ) dt
= f
( z ( ? ( t) ) ) z (?( t ) ) ? ( t ) )
= , u
( z ( ? ( t) ) ) i v( z ( ?(t) ) ) - ,x ( ?( t ) ) i y ( ? ( t) ) -? ( t) d t
=[ u
( z ( ? ( t) ) x ( ? ( t) ) ? ( t) d t v
( z ( ? ( t) ) ) y ( ?( t ) ) ? ( t ) d t- i, u
( z ( ?( t) ) ) y ( ?(t) ) ? ( t) d t
+ v
( z ( ?(t) ) ) x ( ?( t ) ) ? ( t ) d t- ……( ii )
eq
n
(i) & (ii) are equal
by change of variable of real integral
ls o , ? x ( t )
dt ? y ( t )
???? y c ha n ge of va r ia b l e
Put ?(t) = r When
t =c then ?(c)=a & t =d then ?(d)=b
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
? f
= f
ALTERNATE
Let C 1 : z = z(t) , a t b
C 2 : w= w(t) , c t d
Suppose C 1 & C 2 be smoothly equivalent Curve,
? ? ,c, d - ,a , b - s. t.
?( c) = a , ?( d ) b & z ( ?( t) w ( t)
Consider,
c
f = f
(w)dw = f
( w ( t) ) w ( t) d t
= f
( z ( ? ( t) ) ) z (?( t ) ) ) ? ( t) d t p ut ?( t ) p
= f
( z ( p ) ) z (p) d p ? ( t) d t d p
c
f (z)dz
Def
n
. let C b e t he cu r v e giv en b y z z ( t) a t b then
C is d efine d b y z ( b a t) a t b .
(intuitively ?????????? ?????? , ) C has the sa me p o in ts se t of C , traced in th e rev erse d ir
n
.
ote C ler aly , C ( b ) C ( a)
C( a) C ( b )
F ur ther, ( c) ( t ) C ( b a t )
T heor em . Pro v e that ? f
? f
Proof : Consider,
c f f
( z ( b a t ) ) z (b a t) d t
, , * u
( z ( b a t ) )
i v( z ( b a t ) ) +* x ( b a t) ) iy ( b a t ) ) +--dt
, , u
( z ( b a t ) ) x ( b a t)
i v( z ( b a t ) ) x ( b a t ) ) i u ( b a t) y ( b a t) v ( z ( b a t ) y ( b a t) --d t
Put b a t w limit when t = a , w = b
d t d w t = b , w = a
dw d t
? , u( z ( w ) x ( w ) i v ( z ( w ) ) x ( w ) i u ( z ( w ) ) y ( w ) v ( z ( w ) ) y ( w ) - dw
f
( z ( w ) ) z ( w ) d w c f ( w ) d w c f
Alternate
c f = f
( z ( b a t) ) z ( b a t) ( ) d t
Put b a t w
d t d w
f
( z ( t) ) z (t )dt by the property of finite integral
= - c f alternate on remaing portion paper
E x 3 /p g S upp ose f( z ) let C b e smooth c ur v e , then c f (z)dz =?
Sol
n
c f (z)dz= f
(z)dz = f
( z ) z ( t ) d t z ( t) d t
= , z ( t ) -
z ( b ) z ( a)
NOTE let ? ?? b e comp lex num b er then ? « ß if |? | | ?? |
Thm 4.9 Let G(t) be a continuous complex fun
n
of t, then
G
( t) d t « | G |
dt ( absolute value inequality , G
( t) d t )
Proof: let G
(t)dt =Re
i R …… . ( i)
Then, e
G
(t)dt =e
i R e
i R … …. ( ii )
If G
?? (t)dt =0, then above
inequality is obviously true. If not
we write its value in polar form, say
eq
n
(i)
? | G
?? (t)dt| = R
Page 5
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( t
i t
) ( t it ) dt
? ( t
i t
i t
t
) dt
i ? t
dt
i 4
t
5
i
i
3
Ex.1\Pg.46 Evaluate , where f (z) =x
2
+iy
2
and C is given by
z ( t) t i t t
Sol
n
:- z ( t) t it z ( t ) i
But z = x+iy
x =t , y=t
? f ( z ) ( x(t ) )
2
+i (y(t))
2
=t
2
+ i t
2
? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( t
i t
) ( i ) dt
i ? t
dt
i 4
t
3
5
3
i
NOTE: Curve in Que 2 & 3 both are smoothly equivalent by thm 4.5
Qu e 3 ? f
w her e f ( z ) z
C is g iven by z ( t ) sin t i cos t , t
S oltio n f ( z ) z
x iy x i y
x
y
x
x
y
y
x
y
z(t) = sin t + i cost z ( t) cos t – i sin t
but z = x +i y
? x sin t , y = cos t
? x
2
+ y
2
=1
f(z(t)) = sin t –i cos t
? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( sin t i cost ) ( cost i sin t ) dt
? ( sin t cost i ) dt
Ex pg E v aluate ? f
w her e f ( z ) z
C is g iven by z ( t ) R cos t i R sin t , t , R
S olu ti on f ( z ) z
x i y
x i y x
y
z ( t ) R cos t i R sin t z
( t ) R sin t i R cos t
But, z =x+iy
? ( ( ) ) R cos R
R sin R
cos R
sin R
? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
?
cos i sin R
( R sin t i R cos t ) dt
? ( C ost iS in t ) ( S in t iCost ) dt
? ( sin t cos t i sin
t i cos
t cos t sin t ) dt
? ( sin
t cos
t ) dt
i ? . dt
i ( t )
i
T he cu r v e i s sai d to b e smooth iff z ( t) , excep t a t a f in it e n o. of p oi nt s.
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
i. e. iff x ( t) y ( t) ,
Smoothly equivalent Curve Curve C 1 g iven b y z z ( t) a t b is s . t . b . smo ot hly equivalent to Curve C 2 given
b y w w ( t) c t d . If there exist a on e t o one cont in uou sly d iff ma p p in g
? ,c , d - ,a , b - s. t .
(i) ? ( c) a , ? ( d ) b , ? ( t ) = ? t increasing
(ii) z ( ?( t) ) w ( t )
Que. 1 Prove that a smoothly equivalent curve is an equivalence relation.
Reflexive C given by z = z(t) is smoothly equivalent to itself.
S in ce ? can b e t aken a s i d en ti ty f un
c
.
Symmetric let C 1 be smoothly equivalent to C 2.
Let ? ,c, d - ,a , b - b e one to one sat isfy in g cod
n
?( c) a ?( d ) b & z ( ?( t ) ) w ( t) then
?
1
,a, b - ,c, d - w ill g iv es C 2 which is also smoothly equivalent to C 1
Transitive let C 1 : z=z(t) , a t b
C 2 w w ( t) c t d
C 3 r r ( t) e t f
If C 1 be smoothly equivalent to C 2
? ? a one to one map p i ng ? 1 ,c, d - ,a , b - s. t .
? 1 ( c) a, ? 1 ( d ) b z ( ? 1(t))=w(t)
and C 2 be smoothly equivalent to C 3 then
? 2 ( e ) c , ? 2 ( f ) d w ( ? 2(t))=r(t)
let ? ? 1 ° ? 2 then ? i s on e t o one such that
? ,e, f - ,a , b -
?( e) ( ? 1 ° ? 2 ) ( e) ? 1 (? 2 ( e) ) ? 1(c)=a
?( f ) ( ? 1 ° ? 2 ) ( f ) ? 1 (? 2 ( f ) ) ? 1(d)=b
also, z ( ?( t) ) z ( ( ? 1 ° ? 2)(t))
z ( ? 1 (? 2 ( t) ) ) w ( ? 2(t)) = r(t)
Therefore, C 1 is smoothly equivalent to C 3
Hence, a smoothly equivalent curve is an equivalence relation.
T hm . If C
C
are smo othly eq uiv ale nt cu r v e , then ? f
c
? f
c
Proof let C 1 :z = z(t) , a t b
C 2 :w = w(t) , c t d
Let C 1 & C 2 be smoothly equivalent, therefore there exists a mapping
? ,c, d - ,a , b - s. t .
?( c) a , ?( d ) b z ( ?( t ) ) w ( t)
Consider,
? ? f
? f ( z ) dz
? f ( z ( t ) ) z
( t ) dt
= , u
( z ( t) ) iv ( z ( t) ) -,x ( t ) iy ( t) -d t
= , u
( z ( t) ) x ( t ) d t v
( z ( t) ) y ( t) d t- i, u
( z ( t) ) y ( t ) d t v
( z ( t) ) x ( t ) d t …… …( i)
Consider,
? ? f
? f ( w ) dw
? f ( w ( t ) ) w
( t ) dt
= f
( z ( ? ( t) ) ) z (?( t ) ) ? ( t ) )
= , u
( z ( ? ( t) ) ) i v( z ( ?(t) ) ) - ,x ( ?( t ) ) i y ( ? ( t) ) -? ( t) d t
=[ u
( z ( ? ( t) ) x ( ? ( t) ) ? ( t) d t v
( z ( ? ( t) ) ) y ( ?( t ) ) ? ( t ) d t- i, u
( z ( ?( t) ) ) y ( ?(t) ) ? ( t) d t
+ v
( z ( ?(t) ) ) x ( ?( t ) ) ? ( t ) d t- ……( ii )
eq
n
(i) & (ii) are equal
by change of variable of real integral
ls o , ? x ( t )
dt ? y ( t )
???? y c ha n ge of va r ia b l e
Put ?(t) = r When
t =c then ?(c)=a & t =d then ?(d)=b
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
? f
= f
ALTERNATE
Let C 1 : z = z(t) , a t b
C 2 : w= w(t) , c t d
Suppose C 1 & C 2 be smoothly equivalent Curve,
? ? ,c, d - ,a , b - s. t.
?( c) = a , ?( d ) b & z ( ?( t) w ( t)
Consider,
c
f = f
(w)dw = f
( w ( t) ) w ( t) d t
= f
( z ( ? ( t) ) ) z (?( t ) ) ) ? ( t) d t p ut ?( t ) p
= f
( z ( p ) ) z (p) d p ? ( t) d t d p
c
f (z)dz
Def
n
. let C b e t he cu r v e giv en b y z z ( t) a t b then
C is d efine d b y z ( b a t) a t b .
(intuitively ?????????? ?????? , ) C has the sa me p o in ts se t of C , traced in th e rev erse d ir
n
.
ote C ler aly , C ( b ) C ( a)
C( a) C ( b )
F ur ther, ( c) ( t ) C ( b a t )
T heor em . Pro v e that ? f
? f
Proof : Consider,
c f f
( z ( b a t ) ) z (b a t) d t
, , * u
( z ( b a t ) )
i v( z ( b a t ) ) +* x ( b a t) ) iy ( b a t ) ) +--dt
, , u
( z ( b a t ) ) x ( b a t)
i v( z ( b a t ) ) x ( b a t ) ) i u ( b a t) y ( b a t) v ( z ( b a t ) y ( b a t) --d t
Put b a t w limit when t = a , w = b
d t d w t = b , w = a
dw d t
? , u( z ( w ) x ( w ) i v ( z ( w ) ) x ( w ) i u ( z ( w ) ) y ( w ) v ( z ( w ) ) y ( w ) - dw
f
( z ( w ) ) z ( w ) d w c f ( w ) d w c f
Alternate
c f = f
( z ( b a t) ) z ( b a t) ( ) d t
Put b a t w
d t d w
f
( z ( t) ) z (t )dt by the property of finite integral
= - c f alternate on remaing portion paper
E x 3 /p g S upp ose f( z ) let C b e smooth c ur v e , then c f (z)dz =?
Sol
n
c f (z)dz= f
(z)dz = f
( z ) z ( t ) d t z ( t) d t
= , z ( t ) -
z ( b ) z ( a)
NOTE let ? ?? b e comp lex num b er then ? « ß if |? | | ?? |
Thm 4.9 Let G(t) be a continuous complex fun
n
of t, then
G
( t) d t « | G |
dt ( absolute value inequality , G
( t) d t )
Proof: let G
(t)dt =Re
i R …… . ( i)
Then, e
G
(t)dt =e
i R e
i R … …. ( ii )
If G
?? (t)dt =0, then above
inequality is obviously true. If not
we write its value in polar form, say
eq
n
(i)
? | G
?? (t)dt| = R
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
Let e
i
G(t) = A(t) + i B(t)
Where A(t) & B(t) are real value fun
n
of t.
e
G
(t)dt = (t)+i ( t) d t …. ( ii i )
? f r om eq
n
(ii) & (iii)
R= (t)+i (t)dt
(t)dt =R , (t)dt= 0
Now, R= (t)dt = r ea l ( e
G
(t))dt
| e
G
( t) | d t Re( z ) | z|
= | e
| | G
(t)|dt
= | G
(t)| dt
R | G
( t) | d t ……. (iv)
From eq
n
( i) | G
(t)dt| = |Re
i |
| G
(t)dt| =R …… …. . ( v )
? from eq
n
(iv) & (v)
| G
(t)dt |< | G |
dt
ALTERNATE
| G
(t)dt |< | G |
dt
By polar form, also G
(t)dt =R e
i
| G
(t)dt | =R & e
G
(t)dt=R
Taking the real part of both side of the above
R = Re (R) = Re( e
G
(t)dt)
= Re * e
G
(t)}dt
| e
G
(t)|dt using R e( z ) | z |
= | G
(t)|dt | e
i | = 1
Hence, desired inequality hold.
. L F or mu la/ L I ne q ulity Suppose that C is a (smooth) curve of length L, that f is continuous
on C, an d that , f ( i. e . | f | ) thr oug hou t C, then c f ( z ) d z L i. e . | c f(z)dz| <ML (M is the
upper bd of f)
Proof: let the curve C be given by z = z(t) =x(t)+i y(t)
? | f
(z)dz| = | f
( z ( t) ) z (t) d t|
| f
( z ( t) ) z ( t) | d t
= | f
( z ( t) ) | | z ( t ) | d t |
z ( t) | d t
=M |
x ( t) i y ( t) | d t v *x ( t ) +
* y ( t )
+
dt
=M v *
+
*
+
dt =M
dt
=M ds
=ML (where L = ds
= length of the Curve C or L= v * x ( t ) +
* y ( t )
+
dt
? | f
( z ) d z | L
Example/pg 48 let C be the unit circle and suppose f <<1 on C, then M=1, L=2 (length of the circle when
r ) an d c f(z)dz <<2 i. e. | c f ( z ) d z |
Sol
n
Here, M=1
L =length of the curve C = 2
y L f or mu la
| c f ( z ) d z | ? 2 =2
Read More
|
27 docs|150 tests
|
FAQs on Line Integral - Topic-wise Tests & Solved Examples for Mathematics
| 1. What is a line integral in mathematics? |  |
A line integral is a mathematical concept that calculates the cumulative effect of a vector field along a given curve. It measures the total work done or the total accumulated value of a quantity as one moves along the curve.
| 2. How is a line integral calculated? | |
To calculate a line integral, one needs to divide the given curve into small segments and approximate the integral over each segment. The approximation is improved by taking smaller segments. The line integral is then obtained by summing up the approximations for all segments.
| 3. What is the significance of line integrals in physics? | |
Line integrals play a crucial role in physics, particularly in the field of electromagnetism. They are used to calculate quantities such as work done by a force along a path, electric or magnetic field strength, and circulation of a vector field. Line integrals are essential in understanding the behavior of fields and forces in physical systems.
| 4. Can line integrals be negative? | |
Yes, line integrals can be negative. The value of a line integral depends on the orientation of the curve and the vector field being integrated. If the curve and the vector field have opposite directions, the line integral can result in a negative value. It signifies that the work done or the accumulated quantity is in the opposite direction of the curve.
| 5. What are some real-life applications of line integrals? | |
Line integrals find applications in various real-life scenarios. They are used in physics to calculate the work done by a force in moving an object along a curved path. In engineering, line integrals are used to analyze fluid flow, such as calculating the circulation of a fluid in a closed loop. They also have applications in computer graphics, where they are used to render three-dimensional objects by simulating the movement of light along curves.
Related Exams
About this Document
|
1.8K Views |
|
4.71/5 Rating |
|
Dec 18, 2024 Last updated |
Document Description: Line Integral for Mathematics 2024 is part of Topic-wise Tests & Solved Examples for Mathematics preparation.
The notes and questions for Line Integral have been prepared according to the Mathematics exam syllabus. Information about Line Integral covers topics
like and Line Integral Example, for Mathematics 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Line Integral.
Introduction of Line Integral in English is available as part of our Topic-wise Tests & Solved Examples for Mathematics
for Mathematics & Line Integral in Hindi for Topic-wise Tests & Solved Examples for Mathematics course.
Download more important topics related with notes, lectures and mock test series for Mathematics
Exam by signing up for free. Mathematics: Line Integral | Topic-wise Tests & Solved Examples for Mathematics
Description
Full syllabus notes, lecture & questions for Line Integral | Topic-wise Tests & Solved Examples for Mathematics - Mathematics | Plus excerises question with solution to help you revise complete syllabus for Topic-wise Tests & Solved Examples for Mathematics | Best notes, free PDF download
Information about Line Integral
In this doc you can find the meaning of Line Integral defined & explained in the simplest way possible. Besides explaining types of
Line Integral theory, EduRev gives you an ample number of questions to practice Line Integral tests, examples and also practice Mathematics
tests
|
27 docs|150 tests
|
Download as PDF
|
Explore Courses for Mathematics exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Extra Questions
,Previous Year Questions with Solutions
,mock tests for examination
,Free
,Line Integral | Topic-wise Tests & Solved Examples for Mathematics
,Sample Paper
,shortcuts and tricks
,video lectures
,MCQs
,past year papers
,Viva Questions
,Exam
,Objective type Questions
,Semester Notes
,ppt
,Summary
,Line Integral | Topic-wise Tests & Solved Examples for Mathematics
,study material
,Important questions
,practice quizzes
,Line Integral | Topic-wise Tests & Solved Examples for Mathematics
;
Additional Information about Line Integral for Mathematics Preparation
Line Integral Free PDF Download
The Line Integral is an invaluable resource that delves deep into the core of the Mathematics exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Line Integral now and kickstart your journey towards success in the Mathematics exam.
Importance of Line Integral
The importance of Line Integral cannot be overstated, especially for Mathematics aspirants.
This document holds the key to success in the Mathematics exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Line Integral Notes
Line Integral Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Line Integral.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Line Integral Notes on EduRev are your ultimate resource for success.
Line Integral Mathematics Questions
The "Line Integral Mathematics Questions" guide is a valuable resource for all aspiring students preparing for the
Mathematics exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Line Integral on the App
Students of Mathematics can study Line Integral alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Line Integral,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Line Integral is prepared as per the latest Mathematics syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup