LINE INTEGRAL IN A SCALAR FIELD
MOTIVATION
A rescue team follows a path in a danger area where for each
position the degree of radiation is defined. Compute the total
amount of radiation gathered by the rescue team along the path.
RESCUE
BASE
SCALAR FIELD
Let P be a planar contiguous area with a function f (x, y )
defined on it. We call f (x, y ) a scalar field defined over M
Similarly, for a 3D area V, we define a scalar field as a
function f (x, y, z ) with the domain V.
temperature, radiation, moisture, mass, ...
f ( x, y )
[x, y ]
P
V
f ( x, y , z )
[x, y, z ]
Let a scalar field f (x,y) be defined in a planar area M.
Let L be a regular curve defined by the parametric equations
x =ϕ t  y =ψ t , t ∈a,b
Define a partition of T:
a = a 0 < a1 < L < a n
Put
L0 = [ϕ (a0 ),ψ (a0 )], L1 = [ϕ (a1 ),ψ (a1 )],K , Ln = [ϕ (an ),ψ (an )]
Choose points ξ1 , ξ 2 ,K, ξ n such that
a0 < ξ1 < a1 < ξ 2 L < ξ n < an
Put N (a0 , a1 , K, an ) = max{ Li Li −1 }
0<i ≤ n
N (a 0 , a1 , K , a n ) is called the norm of the partition.
a=a0
a1 a 2 a3
ξ1 ξ2
f (ϕ (ξ 1 ),ψ (ξ 1 ))
L0
an-1
...
ξ3
L1
an=b
ξn
Ln
f (ϕ (ξ 2 ),ψ (ξ 2 ))
L2
f (ϕ (ξ 3 ),ψ (ξ 3 ))
...
L3
Ln-1
M
f (ϕ (ξ n ),ψ (ξ n ))
We can define the following integral sum
L = L0 L1 f (ϕ (ξ1 ),ψ (ξ1 )) + L1 L2 f (ϕ (ξ 2 ),ψ (ξ 2 )) + L + Ln −1 Ln f (ϕ (ξ n ),ψ (ξ n ))
If, for N (a0 , a1 , K , an ) → 0 , we have L → L0 for some finite L0,
we say that L0 is the line (or curvilinear) integral of the scalar field
f (x,y) over L (line integral of the first type).
Formally, we write
L0 = ∫ f (x, y ) ds
L
L0 = ∫ f (x, y ) ds is independent of the way the partition of
L
[a,b] is constructed and the numbers ξ1 , ξ 2 , K , ξ n chosen.
If f(x,y) is continuous on a regular curve L given by the
parametric equations x =ϕ t  y =ψ t , t ∈a,b then the following
formula can be used
∫
L
or
∫
L
b
f (x, y ) ds = ∫ f (ϕ (t ),ψ (t )) ϕ '2 (t ) + ψ '2 (t ) dt
a
b
f (x, y ) ds = ∫ f ( x, g ( x )) 1 + g '2 ( x ) dt
a
if L is given by the explicit equation y=g (x).
For a 3D-area V, a scalar field f (x,y,z), and a 3D-curve L the
line integral is defined in much the same way. If f (x,y,z) is
continuous on L given by x =ϕ t  y =ψ t  z =τ t , t ∈a,b
we can use the formula
∫
L
b
f (x, y, z ) ds = ∫ f (ϕ (t ),ψ (t ), τ (t )) ϕ '2 (t ) + ψ '2 (t ) + τ '2 (t ) dt
a
If we recall the way the length of a curve is defined, we see
that ds actually denotes the differential of the length of L.
EXAMPLE
ds
Calculate ∫ x 2 + y 2 + z 2
C
where C is the first turn of the helix
x = a cos t , y = a sin t , z = bt , t ∈ [0, π ]
In the plot below, we have a=2 and b=0.3
1
x2 + y2 + z 2
[x,y,z]
2π
ds
∫0 x 2 + y 2 + z 2 =
2π
∫
0
a 2 sin 2 t + a 2 cos 2 t + b 2
dt =
2
2
2
2 2
a cos t + sin t + b t
2π
(
a2 + b2
a 2 + b2
=∫ 2
dt =
2 2
2
a
+
b
t
a
0
a a2 + b
=
a 2b
2 bπ
2 a
∫
0
)
2π
 bt

=
=
u
∫0  (bt )2   a  =
1+ 
2
a 

dt
2 bπ
du
a 2 + b2
[arctan u ]0 a =
=
2
1+ u
ab
a 2 + b2
2bπ
=
arctan
ab
a
EXAMPLE
Find the centre of mass of the first half-arch of the
cycloid
x = a(t − sin t ), y = a(1 − cos t ), 0 ≤ t ≤ π
if its specific mass is constant.
The centre of mass of a curve is defined in physics as the point
T=[tx,ty] that has the following property: if the entire mass of the
curve was concentrated at T, its torque with respect to axes x, y
would be equal to the torque with respect to axes x, y of all the
points added up. This means that
tx =
∫ x ds
C
∫ ds
C
, ty =
∫ y ds
C
∫ ds
C
π
∫ ds = ∫
π
(a − a cos t )2 + a 2 sin 2 t dt =
2a ∫ 1 − cos t dt =
0
C
0
π
= 2a ∫
0
π
π
2 1 − cos t
t
t

dt = 2 2a ∫ sin dt = −4 2a cos  = 4 2a
2
2
2 0

0
π
2
2
(
)
x
ds
=
a
2
1
−
cos
t
t
−
sin
t
dt
=
a
2 (I1 − I 2 )
∫
∫
C
π
0
π
where I1 = ∫ t 1 − cos t dt and I 2 = ∫ sin t 1 − cos t dt
0 π
0
2
2
(
)
y
ds
=
a
2
1
−
cos
t
1
−
cos
t
dt
=
a
2 (I 3 − I 4 )
∫
∫
C
π
0
π
where I 3 = ∫ 1 − cos t dt and I 4 = ∫ cos t 1 − cos t dt
0
0
π
π
π
2
t
1 − cos t dt = 2 ∫ sin dt = 4 ∫ sin u du = 4
2
0
0
I3 = ∫
0
π
π
π
t
t

I1 = ∫ t 1 − cos t dt = − 4t cos  + 4 ∫ cos dt = 0 + 8 = 8
2 0
2

0
0
π
π
t
t

I 2 = ∫ sin t 1 − cos t dt = 2∫ sin t {sin
dt
=
=
u

=
}
2
2

0
0
π
2
π
2
1
= 4 ∫ sin 2u sin u du = 8 ∫ sin 2 u cos u dt = {sin u = v} = 8∫ t 2 dt =
0
π
0
0
π
t
t

I 4 = ∫ cos t 1 − cos t dt = 2∫ cos t sin dt =  = u  =
2
2

0
0
8
3
π
π
2
2
= 4 ∫ cos 2u sin u du = 4 ∫ cos 2 u sin u − cos3 u dt =
0
π
0
2
= 4 ∫ cos 2 u sin u − cos u + cos u sin 2 u dt =
0
3 1
3 1
t 
t 
8
8 −4
2
= 4  − 4[sin t ]0 + 4   = − 4 + =
3
3
 3 0
 3 0 3
∫ x ds = a
π
2
C
2
=
y
ds
a
∫
C
16a 2 2
2 (I1 − I 2 ) =
3
16a 2 2
2 (I 3 − I 4 ) =
3
16a 2 2
4a
3
tx = t y =
=
3
4 2a
LINE INTEGRAL IN A VECTOR FIELD
MOTIVATION
A ship sails from an island to another one along a fixed route.
Knowing all the sea currents, how much fuel will be needed ?
VECTOR FIELD
Let P be a planar contiguous area with a vector function
f ( x, y ) = f1 ( x, y )i + f1 ( x, y ) j defined for each [x, y ]∈ P.
We say that f ( x, y ) is a vector field on P. Similarly,
f ( x, y, z ) = f1 ( x, y, z )i + f 2 ( x, y, z ) j + f 3 ( x, y, z )k is a vector field
in a 3D area V.
water flow, electromagnetic field, ...
V
P
f ( x, y )
[x, y ]
f ( x, y , z )
[x, y, z ]
Let a vector field f ( x, y ) be defined on a planar area M.
Let L be an oriented regular curve defined by the parametric
equations x =ϕ t  y =ψ t , t ∈a,b


Define a partition of T:
a = a 0 < a1 < L < a n
Put L0 = [ϕ (a0 ),ψ (a0 )], L1 = [ϕ (a1 ),ψ (a1 )],K, Ln = [ϕ (an ),ψ (an )]
Choose points ξ1 , ξ 2 ,K, ξ n such that
a 0 < ψ 1 < a1 < ψ 2 L < ψ n < a n
Put
{
N (a0 , a1 , K, an ) = max Li Li −1
0<i ≤ n
}
N (a 0 , a1 , K , a n ) is called the norm of the partition.
a=a0
a1 a 2 a3
ξ1 ξ2
f (ϕ (ξ1 ),ψ (ξ1 ))
ξ3
L1
an=b
ξn
Ln
f (ϕ (ξ 2 ),ψ (ξ 2 ))
L2
L0
an-1
...
f (ϕ (ξ 3 ),ψ (ξ 3 ))
...
L3
Ln-1
M
f (ϕ (ξ n ),ψ (ξ n ))
Define the following integral sum
n
L = ∑ Li −1 Li f (ϕ (ξ i ),ψ (ξ i ))
i =1
If, for N (a0 , a1 , K , an ) → 0 , we have L → L0 for some finite L0,
we say that L0 is the line (or curvilinear) integral of the vector
field f ( x, y ) over L (line integral of the second type).
Symbolically, we write
L0 = ∫ f (x, y ) ds
or
L
L0 = ∫ f1 ( x, y ) dx + f 2 ( x, y )dy
L
L0 = ∫ f ( x, y ) ds = ∫ f1 (x, y )dx + f 2 ( x, y )dy
L
L
is independent of the way we construct the partition of [a,b] and
choose the numbers ξ1 , ξ 2 , K , ξ n
To calculate the line integral we can use the formula
b
∫ f (x, y ) ds = ε ∫ f (ϕ (t ),ψ (t ))ϕ ' (t ) + f (ϕ (t ),ψ (t ))ψ ' (t ) dt
1
L
2
a
where ε is 1 if the curve is oriented in correspondence with
its parametric equations otherwise it is equal to –1.
For a 3D-area V, a vector field f ( x, y, z ) and a 3D-curve L
the line integral is defined in much the same way with the
following formula for calculating a 3D line integral:
b
∫ f ( x, y, z ) ds = ε ∫ ( f (ϕ ,ψ ,τ )ϕ '+ f (ϕ ,ψ ,τ )ψ '+ f (ϕ ,ψ ,τ )τ ' ) dt
1
L
2
3
a
ds
Line integral of the second type clearly depends on the way the
curve is oriented and ds actually denotes the differential of
the tangent vector field along L.
EXAMPLE
Calculate
∫ ( y − z ) dx + (z − x ) dy + (x − y ) dz
L
where L is the first turn of the helix oriented in correspondence
with the parametric equations
x = a cos t , y = a sin t , z = bt , t ∈ [0, π ]
1
x2 + y2 + z 2
[x,y,z]
a=2 and b=0.3
∫ ( y − z ) dx + (z − x ) dy + (x − y ) dz =
L
= ∫ (a sin t − bt )(− a sin t ) + (bt − a cos t ) a cos t + (a cos t − a sin t ) b dt =
L
2π
= ∫ − a 2 sin 2 t + abt sin t + abt cos t − a 2 cos 2 t + ab cos t − ab sin t dt =
0
2π
2π
2π
2π
 2π 2
= ab ∫ t sin t dt + ∫ t cos t dt − ∫ cos t dt − ∫ sin t dt  − ∫ a dt =
0
0
0
0
 0
2π
2π

= ab  ∫ t sin t dt + ∫ t cos t dt  − 2πa 2 =
0
0

2π
2π


2π
2π
= ab− [t cos t ]0 + ∫ cos t dt + [t sin t ]0 − ∫ sin t dt  − 2πa 2 = −2πa(a + b )
0
0


EXAMPLE
Calculate
∫ ydx + zdy + xdz
where C is a circle given by
C
the parametric equations
x = R cos α cos t , y = R cos α sin t , z = R sin α , α = const
oriented in correspondence with them.
z
α
x
R
y
∫ ydx + zdy + xdz =
C
2π
= ∫ R cos α sin tR cos α (− sin t ) + R sin αR cos α cos t + R cos α cos t 0 dt =
0
2π
= R 2 ∫ − cos 2 α sin 2 t + sin α cos α cos t dt =
0
2π
2π
0
0
= − R 2 cos 2 α ∫ sin 2 t dt + R 2 sin α cos α ∫ cos t dt =
2π
1 − cos 2t
dt = −π R 2 cos 2 α
2
0
= − R 2 cos 2 α ∫
EXAMPLE
Determine the work done by the elasticity force if the vector
of this force at each point is directed towards the origin its
module being proportionate to the distance from the origin of
the point and if the application point of the force moves
anticlockwise along the quarter of the ellipse x 2 y 2
+ 2 =1
2
a
b
that lies in the first quadrant.
b
a
W = ∫ f1 ( x, y )dx + f 2 ( x, y )dy where f1 ( x, y ) = −kx, f 2 ( x, y ) = −ky
L
and L is given by the parametric equations
[
x = a cos t , y = b sin t , t ∈ 0, π
π
2
]
2
W = ∫ − ka cos t (− a sin t ) − kb sin tb cos t dt =
0
(
= k a2 − b2
π
) ∫ sin t cos t dt = {sin t = u} =
2
0
1
(
2
2


u
k
a
−
b
= k a2 − b2   =
2
 2 0
(
)
2
)
tends to zero as
the ellipse tends
to circle