¿Cuál es el camino más corto para un cable eléctrico?
Planta eléctrica
ciudad
planta eléctrica
Ejemplo típico que motiva la necesidad de integrales de contorno
Integrales de contorno o de camino
Secciones cónicas
Círculo Elipse Parábola Hipérbola
Ecuación general de una sección cónica:Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
Longitud de una curva
• z’(t) = x’(t) + iy’(t) • Si x’(t) e y’(t) son continuas en el intervalo a <= t <= b entonces: C es un arco diferenciable y su longitud es:
b
a
dttzL |)('|
Si z’(t) no es 0 en ningún punto de a < t < b entonces podemos definir un vector tangente:
|)('|
)('
tz
tzT Y decimos que z(t) es un arco suave.
Contorno = arco suave a trozos
2ty
2txtz )(')(')('
y
x
C: z(t)
Ejemplo: la longitud de camino es independiente de la parametrización
a
b
rx= r cos(t)
y= r sin(t)
0 t π/2
2|)(sin')(cos'|
2/
0
2/
0
2/
0
2/
0
rdtrdterdt
dt
derdttirtrL it
it
1|| ititit eeezzz
x= r cos(2)
y= r sin(2)
0 π
222|)2(sin')2(cos'|
00
2
0
2
0
r
dr
der
dd
derdirrL i
i
Dos parametrizaciones distintas
2
2
and where,
11
aty(t) tx(t) iyxz
xaxy
1
1
21
1
1
1
1
1
)2(1|21||)(')('||)('| dtatdtatidttiytxdttzL
sin Substitute iw
atw 2 Substitute a
dwwa
L2
0
211
dwwI 21Let
didiI 22 cos)cos(sin1
a
-1x
y
2
2sin1cos2
2
2sin
2
iI
)(sin Substitute 1 iw
2
)(1)(2)(sin
2
21 iwiw
iwi
I
1
2412ln4122
1
2
0
21ln12
1
2a
0
2w11
L
2
2
aaaaa
aw
wwwww
a
dwa
21ln)(1-sinh)(1-sin :identity following theUse
wwwiwi
dwwwwwwI 222 111ln2
1
2
)(1)(2)(sin
2
21 iwiw
iwi
I
a
-1x
1222 )5.0(5.0
21 a
La
62.125.10.5)5.0(5.0
48152ln2
15
2
2
2
1
2
1.4121
,1When
22
2
a
.. L
a
a
0.5
axy 2'
Circulation and Net Flux• Let T and N denote the unit tangent vector and the
unit normal vector to a positively oriented simple closed contour C. When we interpret the complex function f(z) as a vector, the line integrals
(6)
(7)
have special interpretations.
C C
dyvxdusdf T
C C
vdyxdusdf N
• The line integral (6) is called the circulation around C and (7) is called the net flux across C. Note that
and socirculation = (8)
net flex = (9)
C C
CC
zdzfdyixdivu
sdfisdf
)())((
NT ..
C zdzf )(Re
C zdzf )(Im
Given the flow f(z) = (1 + i)z, compute the circulation around and the net flux across the circle C: |z| = 1.
Solution
2 ,2ncirculatio
)1(2)1()1()(
20,)( and )1()( Since2
0
2
0
fluxnet
idtidtieeidzzf
tetzzizf
itit
C
it
The complex function where k = a + ib and z1 are complex numbers, gives rise to a flow in the domain z z1. If C is a simple closed contour containing z = z1 in its interior, then we have
The circulation around C is 2b and the net flux across C is 2a. If z1 were in the exterior of C both of them would be zero.
)/()( 1zzkzf
)(2)(1
ibaizdzziba
zdzfC C
Note that when k is real, the circulation around C is zero but the net flux across C is 2k. The complex number z1 is called a source when k > 0 and is a sink when k < 0.
Evaluate
where C is the circle |z| = 1.
SolutionThis integrand is not analytic at z = 0, −4 but only z = 0 lies within C. Since
We get z0 = 0, n = 2, f(z) = (z + 1)/(z + 4), f (z) = −6/(z + 4)3. By (6):
33441
4
1
zzz
zz
z
zdzz
zC
34 4
1
ifi
zdzz
zC 32
3)0(
!22
4
134
Evaluate zdizz
zC
2
3
)(
3
SolutionThough C is not simple, we can think of it is as the union of two simple closed contours C1 and C2 in Fig 18.27.
21
2
3
2
3
22
3
2
3
21
21
)(
3)(
3
)(
3
)(
3
)(
3
II
zdizz
zz
zdz
izz
z
zdizz
zzd
izz
zzd
izz
z
CC
CCC
For I1 : z0 = 0 , f(z) = (z3 + 3)/(z – i)2 :
For I2 : z0 = i, n = 1, f(z) = (z3 + 3)/z, f ’(z) = (2z3 –3 )/z2:
We get
ifizdz
izz
z
IC
6)0(2)(
3
1
2
3
1
)32(2)23(2)(!1
2
)(
3
22
3
2 iiiifi
zdiz
zz
IC
)31(4)32(26)(
3212
3
iiiIIzdizz
zC
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