HUT – DEPARTMENT OF MATH.

APPLIED
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INTERNATIONAL PROGRAM
GIAÛI TÍCH 2

LEC 05: LINE INTEGRALS

STEWART: PAGES 1034 – 1061

• PhD. NGUYEÃN QUOÁC LAÂN (October, 2014)

 INTEGRAL CALCULUS
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Double (2D) Integral ∫∫ f ( x, y )dA

D

Triple (3D) Integral ∫∫∫ f ( x, y, z )dV

E
Highschool :1D Integral
Line Integral along (C) : curve
b

∫ f ( x )dx , [a , b ] ⊂ R 1st kind : ∫ f ( x, y )ds

a (C )

a b
2 nd kind : ∫ P( x, y )dx + Q( x, y )dy
(C )

Surface Integral over (S) : surface.1st kind ∫∫ f ( x, y )dS , 2 nd kind ∫∫ F ⋅ ndS

(S) (σ )
 LINE INTEGRAL ALONG THE CURVE (C): 1st KIND
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(C) : one arc between A , B & ds : Length

A B ⇒ Line Int. ∫ f ( x, y )ds (1st kind or with

2 3 (C )
respect length). Note : A → B ≡ B → A
a b

⎡ 2
[ ]
⎢ y = y ( x ) ⇒ ds = 1 + y ( x ) dx, x = x( y ) ⇒ ds = 1 + x ( y ) dy
/ / 2
[ ]
⎢ 2
⎣ x = x(t ), y = y (t ) ⇒ ds = x (t ) + y (t ) dt
/ /
[ ] [ ]
2
(C) : α ≤ x ≤ β
β ⎡ A→β
2
[ ]
(C) : y = y ( x ) ⇒ I = ∫ f ( x, y ( x )) 1 + y / ( x ) dx; x = x( y )... Even ⎢B → α :
⎣
α
β
β
⎧ x = x(t )
(C) : ⎨
2 2
⇒ I = ∫ f ( x(t ), y (t )) x (t ) + y (t ) dt
/ /
[ ] [ ] ∫ ...ds = ∫ f
⎩ y = y (t ) α
(C ) α
 EXAMPLE
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Evaluate ∫ 2 xds, where C consists of the arc C1 of parabola y = x 2

C
from (0, 0) to (1, 1) followed by vertical line C 2 from (1, 1) to (1, 2)

Answer : C = C1 U C2 ⇒ ∫ fds = ∫ fds + ∫ fds

C C1 C2
1
C1 : y = x 2 , 0 ≤ x ≤ 1 ⇒ I1 = ∫ 2 x 1 + 4 x 2 dx
0

u = 1 + 4 x 2 ⇒ du = 8 xdx ⇒ I1 = (5 5 − 1) 6

⎧ x = 1 2
C2 : ⎨ , 1 ≤ y ≤ 2 ⇒ I 2 = ∫ 2 0 2 + 12 dy = 2
⎩y = y 1
Remark: Line Integral between A & B depends on path AB.
 EXAMPLE
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Important: Find the “good” (suitable) equation to represent C

( )
Evaluate ∫ 2 + x 2 y ds, where C is upper half of circle x 2 + y 2 = 1
C

Parametric equation : y = 1 − x 2
→ Complicated for calculation y / !
⎧ x = cos t
Better (polar!) : ⎨ , t ∈ [0, π ]
⎩ y = sin t
π
( )
I = ∫ 2 + cos 2 t sin t sin 2 t + cos 2 t dt
0
π π 1
2
I = ∫ 2dt + ∫ cos t sin tdt. Change u = cos t → ∫ u du ⇒ I = 2π +
2 2

0 0 −1
3
 LINE INTEGRAL IN SPACE
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(C): space curve x = x(t), y = y(t), z = z(t), a ≤ t ≤ b. Along (C):

[x (t )] + [y (t )] + [z (t )]
b
2 2 2
∫ f ( x, y, z )ds = ∫ f ( x(t ), y(t ), z(t ))
/ / /
dt
C a

Evaluate ∫ y sin zds, where C is x = cos t , y = sin t , z = t , t ∈ [0, 2π ]

C
2π
Answer : ∫ y sin zds = ∫ sin t ⋅ sin t ⋅ sin 2
t + cos 2
t + 1dt
C 0
2π 2π
2 2⎡ 1 ⎤ = 2π
=
2 ∫ [1 − cos 2t ]dt = t
2 ⎢⎣ 2
− sin 2t ⎥⎦
0
0
 LINE INTEGRAL WITH RESPECT TO x AND TO y: 2nd KIND
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Given two functions of two variables

P( x, y ), Q( x, y ) & (C ) from A to B ⇒
⇒ Line Integral of 2 nd kind along (C)
B
A
Δxi from A to B : ∫ P( x, y )dx + Q( x, y )dy
(C )

Distinguish with ∫ f ( x, y )ds : Line integral with respect to arc length

C
A→x=a
[ ]
b
(C) : y = y ( x ), ⇒ ∫ Pdx + Qdy = ∫ P( x, y ( x )) + Q K y / dx
B→ x=b C:A →B a

( )
= → =
( )
b
⎧ x x t A t a
(C) : ⎨ , ⇒ ∫ Pdx + Qdy = ∫ P( x(t ), y (t )) ⋅ x / + Q K ⋅ y / dt
⎩ y = y (t ) B → t = b C a
 EXAMPLE
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Evaluate ∫ y 2 dx + xdy : (a) C = C1 is the line segment from (− 5,−3)

C

to (0,2 ) and (b) : C = C 2 is the arc of x = 4 − y 2 from (− 5,−3) to (0,2 )

⎧ y (− 5) = −3
(a) y = ax + b ⇒ ⎨ ⇒ y = x+2
⎩ y (0 ) = 2
⎧y = x + 2
[ ]
0
C1 : ⎨ ⇒ I1 = ∫ ( x + 2 )2 + x ⋅1 dx
⎩ x : −5 → 0 −5
Result : I1 = − 5 6
⎧x = 4 − y 2
[ ( )]
2
245
(b) C 2 : ⎨ ⇒ I 2 = ∫ y 2 ⋅ (− 2 y ) + 4 − y 2 dy =
⎩ y : −3 → 2 −3
6

The value of a line integral depends on the endpoints and path!

 INDEPENDENCE OF PATH
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C1 Definition : ∫ Pdx + Qdy is INDEPENDENT OF PATH

A
C2
B if ∫ Pdx + Qdy = ∫ Pdx + Qdy for any two path C1, C2
C1 C2
with the same initial point A and terminal point B

Theorem : ∫ Pdx + Qdy is independent of path in D ⇔ ∫ Pdx + Qdy = 0

C
for every closed path C in D ⇔ Py = Qx in D. In this case, if we find
u ( x, y ) : u x = P & u y = Q ⇒ ∫ Pdx + Qdy = [u ]B
A = u ( B ) − u ( A)
A→ B

⎧u x = P ⇒ u = ∫ Pdx = L + C ( y )
⎪
Find u ( x, y ) : ⎨
⎪⎩ y
u = Q ⇒ C /
( y) = L ⇒ C( y) = L
 EXAMPLE
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(
a/ Show that ∫ (3 + 2 xy )dx + x 2 − 3 y 2 dy is independent of path )
b/ Find a function u ( x, y ) : u x = 3 + 2 xy & u y = x 2 − 3 y 2
⎧ =
( )
t
⎪ x e sin t
c/ Evaluate ∫ (3 + 2 xy )dx + x − 3 y dy, C : ⎨
2 2
,t :0 →π
C ⎪⎩ y = e cos t
t

∂P ∂Q
Answer : a/ P = 3 + 2 xy, Q = x − 3 y ⇒ = 2x & = 2 x : OK
2 2
∂y ∂x
b/ u x = 3 + 2 xy. Integrate res. to x : u = ∫ (3 + 2 xy )dx = 3 x + yx 2 + C ( y )

Differentiate u with respect to y : u y = x 2 + C / ( y ) = x 2 − 3 y 2

⇒ C ( y ) = − ∫ 3 y 2 dy = − y 3 + C ⇒ u ( x, y ) = 3 x + x 2 y − y 3

(
c/ t = 0 ⇒ A(0, 1), t = π ⇒ B 0,−eπ ⇒ I = u 0,−eπ − u (0,1) = e3π + 1 ) ( )
 GREEN’S THEOREM
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Consider a simple closed curve C which
bounds the plane region D. Convention:
The positive orientation of the curve C is
counterclockwise direction of C. It means
the region D is always on the left.

Green theorem: C - positively oriented,

simple closed curve; D – bounded by C
⎡ ∂Q ∂P ⎤
∫ P( x, y )dx + Q( x, y )dy = ∫∫ ⎢⎣ ∂x − ∂y ⎥⎦ dA
C D

If C is closed & Line integral along C: difficult → Double Integral

 EXAMPLE
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Evaluate ∫ x 4 dx + xydy, where C is the triangular curve consisting

C
of the line segments from (0, 0) → (1, 0) → (0, 1) → (0, 0)

B
⎡
Green : ∫ x 4 dx + xydy = ∫∫ ⎢
∂ ( xy )
−
∂ x 4
( )⎤⎥ dxdy
C D⎣
∂x ∂y ⎦
1 1− x 1
1 1
=∫ ∫ ydydx = 2 ∫ (1 − x ) dx = 6
2
A
0 0 0

( ) (
Evaluate ∫ 3 y − esin x dx + 7 x + y 4 + 1 dy, C is circle x 2 + y 2 = 9 )
C

I= ∫∫ ⎢
(
⎡ ∂ 7 x + y 4 + 1 ∂ 3 y − esin x
−
) ( )⎤⎥ dxdy =
∫∫ 4dxdy = 4S D = 36π
2 2 ⎢⎣ ∂x ∂y ⎥⎦
x + y ≤9 D