Line Integral

LINE INTEGRAL
Learning outcomes
Students should be able to

 evaluate the line integrals with respect to arc length
 evaluate the line integrals with respect to x, y and z
 relate the line integral and area under the graph of a function
 relate the gradient of a graph and the gradient vector
 sketch a vector field on R2
 determine the gradient vector field of a given multivariable function
 explain the Fundamental Theorem for Line Integrals
Introduction
 The base of a circular fence is given by 𝑥 = cos(𝑡), 𝑦 = sin 𝑡 , 0 ≤ 𝑡 ≤ 2𝜋.

The height of the fence at position (x, y) is given by 𝑓 𝑥, 𝑦 = 4. Sketch the fence on R3.
What is the area of one side of the fence?
- we can use the line integral to calculate this area
Line integrals with respect to arc
length
• Let say you are given a smooth curve C on R3

and a function f defined on C.
• Subdivide C into small sections using distinct
points P = P0, P1, P2, …, Pn = Q.
• Write ∆𝑠𝑘 = arc length between 𝑃𝑘−1 and 𝑃𝑘 .
* * * *
• Choose points Pk ( xk , yk , zk ) between 𝑃𝑘−1 and
𝑃𝑘 .
n
• Then, calculate  f (x , y , z
k 1
*
k
*
k
*
k )sk
Line integral of f n
with respect to s  f ( x, y, z )ds  lim  f ( xk* , yk* , z k* )sk
n 
along C C k 1
Formula
 In R2, let 𝑥 = 𝑥(𝑡) and 𝑦 = 𝑥(𝑡) be the parametric equations defining the smooth curve C
traversed exactly once as t increases from a to b. Then,
𝑏 2 2
𝑑𝑥 𝑑𝑦
න 𝑓 𝑥, 𝑦 𝑑𝑠 = න 𝑓(𝑥 𝑡 , 𝑦 𝑡 ) + 𝑑𝑡
𝑎 𝑑𝑡 𝑑𝑡
𝐶
 Similarly, in R3, let 𝑥 = 𝑥 𝑡 , 𝑦 = 𝑦(𝑡) and 𝑧 = 𝑧(𝑡) be the parametric equations defining
the smooth curve C traversed exactly once as t increases from a to b. Then,
𝑏 2 2 2
𝑑𝑥 𝑑𝑦 𝑑𝑧
න 𝑓 𝑥, 𝑦 𝑑𝑠 = න 𝑓(𝑥 𝑡 , 𝑦(𝑡), 𝑧(𝑡)) + + 𝑑𝑡
𝑎 𝑑𝑡 𝑑𝑡 𝑑𝑡
𝐶
Other notations
 Integrals along C in R2:

𝐫 𝑡 =𝑥 𝑡 𝐢+𝑦 𝑡 𝐣
𝑏 2 2 𝑏
𝑑𝑥 𝑑𝑦
න 𝑓 𝑥, 𝑦 𝑑𝑠 = න 𝑓(𝑥 𝑡 , 𝑦 𝑡 ) + 𝑑𝑡 = න 𝑓 𝐫 𝑡 |𝐫 ′ 𝑡 |𝑑𝑡
𝐶 𝑎 𝑑𝑡 𝑑𝑡 𝑎

𝐫 𝑡 =𝑥 𝑡 𝐢+𝑦 𝑡 𝐣+𝑧 𝑡 𝐤
𝑏 2 2 2 𝑏
𝑑𝑥 𝑑𝑦 𝑑𝑧
න 𝑓 𝑥, 𝑦, 𝑧 𝑑𝑠 = න 𝑓(𝑥 𝑡 , 𝑦 𝑡 , 𝑧(𝑡)) + + 𝑑𝑡 = න 𝑓 𝐫 𝑡 |𝐫 ′ 𝑡 |𝑑𝑡
𝐶 𝑎 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑎
Line integrals for piecewise-smooth
curve
 Suppose now that C is a piecewise-smooth curve; that

is C is a union of a finite number of smooth curves
𝐶1 , 𝐶2 , … , 𝐶𝑛 where the initial point of 𝐶𝑖+1 is the
terminal point of 𝐶𝑖 . Then we define the integral of f
along C as the sum of the integrals of f along each of
the smooth pieces of C:
න 𝑓 𝑥, 𝑦 𝑑𝑠
𝐶
= න 𝑓 𝑥, 𝑦 𝑑𝑠 + න 𝑓 𝑥, 𝑦 𝑑𝑠 + ⋯ + න 𝑓 𝑥, 𝑦 𝑑𝑠
𝐶1 𝐶2 𝐶𝑛
Interpretation of line integrals in 3D
 We can interpret the line integral of a positive

function as an area.
 If 𝑓(𝑥, 𝑦) ≥ 0, then ‫𝑥 𝑓 𝐶׬‬, 𝑦 𝑑𝑠 represents

the area of one side of the “fence” or curtain,
whose base is C and whose height above the
point (x, y) is f(x,y).
Fence problem
 The base of a circular fence is given by

𝑥 = cos(𝑡), 𝑦 = sin 𝑡 , 0 ≤ 𝑡 ≤ 2𝜋. The
height of the fence at position (x, y) is given
by 𝑓 𝑥, 𝑦 = 4. Determine the area of one side
of the fence.
Fence problem
 Solution:
𝑑𝑥 𝑑𝑦
 𝑥 = cos 𝑡 , 𝑑𝑡 = − sin 𝑡 . 𝑦 = sin 𝑡 , 𝑑𝑡 = cos(𝑡).
2𝜋
 ‫ 𝐶׬‬4𝑑𝑠 = ‫׬‬0 4 (− sin 𝑡 )2 +(cos 𝑡 )2 𝑑𝑡 = 4 2𝜋 − 4 0 = 8𝜋.
Example 1
 Evaluate ‫ 𝐶׬‬2 + 𝑥 2 𝑦 𝑑𝑠, where C is the upper half of the circle 𝑥 = cos 𝑡 , 𝑦 = sin 𝑡 , 0 ≤ 𝑡 ≤ 𝜋.
2
 Answer: 2𝜋 + 3
Example 2
 Evaluate ‫ 𝐶׬‬2𝑥 𝑑𝑠 , where C consists of the arc 𝐶1 of the parabola 𝑥 = 𝑡, 𝑦 = 𝑡 2 from (0,
0) to (1, 1) followed by the vertical line segment 𝐶2 𝑥 = 1, 𝑦 = 𝑡 from (1, 1) to (1, 2).
5 5−1
 Answer: + 2
6
Example 3
 Evaluate the line integral ‫𝑠𝑑 𝑧𝑦 𝑒𝑥 𝐶׬‬, where C is the line segment from (0, 0, 0) to (1, 2, 3).
14(𝑒 6 −1)
 Answer: 12
Line integrals of f along C with respect
to x, y and z
n
 f ( x, y ) dx  lim  f ( xk* , yk* )xk

n 
C k 1
n
 f ( x, y ) dy  lim  f ( xk* , yk* )yk

n 
C k 1

n
 f ( x, y, z ) dx  lim  f ( xk* , yk* , z k* )xk

n 
C k 1
n
 f ( x, y, z ) dy  lim  f ( xk* , yk* , z k* )yk

n 
C k 1
n
 f ( x, y, z ) dz  lim  f ( xk* , yk* , z k* )z k

n 
C k 1
Formula
 𝑥 = 𝑥 𝑡 , 𝑦 = 𝑦 𝑡 , 𝑧 = 𝑧 𝑡 → 𝑑𝑥 = 𝑥 ′ 𝑡 𝑑𝑡, 𝑑𝑦 = 𝑦 ′ 𝑡 𝑑𝑡, 𝑑𝑧 = 𝑧 ′ 𝑡 𝑑𝑡
𝑏
 ‫𝑥 𝑓 𝐶׬‬, 𝑦 𝑑𝑥 = ‫ 𝑡 𝑥 𝑓 𝑎׬‬, 𝑦 𝑡 𝑥 ′ 𝑡 𝑑𝑡 ,
𝑏
 ‫𝑥 𝑓 𝐶׬‬, 𝑦 𝑑𝑦 = ‫ 𝑡 𝑥 𝑓 𝑎׬‬, 𝑦 𝑡 𝑦 ′ 𝑡 𝑑𝑡 ,
𝑏
 ‫𝑥 𝑓 𝐶׬‬, 𝑦, 𝑧 𝑑𝑥 = ‫ 𝑡 𝑥 𝑓 𝑎׬‬, 𝑦 𝑡 , 𝑧(𝑡) 𝑥 ′ 𝑡 𝑑𝑡,
𝑏
 ‫𝑥 𝑓 𝐶׬‬, 𝑦, 𝑧 𝑑𝑦 = ‫ 𝑡 𝑥 𝑓 𝑎׬‬, 𝑦 𝑡 , 𝑧 𝑡 𝑦 ′ 𝑡 𝑑𝑡,
𝑏
 ‫𝑥 𝑓 𝐶׬‬, 𝑦, 𝑧 𝑑𝑧 = ‫ 𝑡 𝑥 𝑓 𝑎׬‬, 𝑦 𝑡 , 𝑧 𝑡 𝑧 ′ 𝑡 𝑑𝑡,
Example 4
 Evaluate ‫ 𝑦 𝐶׬‬2 𝑑𝑥 + 𝑥𝑑𝑦 where C is the line segment from (−5, −3) to (0,2).
5
 Answer: − 6
Example 5
 Evaluate ‫ 𝑥𝑑𝑦 𝐶׬‬+ 𝑧𝑑𝑦 + 𝑥𝑑𝑧 where C consists of the line segment 𝐶1 from (2, 0, 0) to
(3, 4, 5) followed by the vertical line segment 𝐶2 from (3, 4, 5) to (3, 4, 0).
 Answer: ‫ 𝑥𝑑𝑦 𝐶׬‬+ 𝑧𝑑𝑦 + 𝑥𝑑𝑧 = 24.5 , ‫ 𝑥𝑑𝑦 𝐶׬‬+ 𝑧𝑑𝑦 + 𝑥𝑑𝑧 = −15.
1 2
 Adding the values of these integrals, we obtain 24.5 – 15 = 9.5

https://www.youtube.com/watch?v=AqcbyjaSQ10
Gradient vector
 If f is a function of two variables x and y, then the gradient of f is the vector function 𝛻𝑓
𝜕𝑓 𝜕𝑓 𝜕𝑓 𝜕𝑓
defined by 𝛻𝑓 𝑥, 𝑦 = , = 𝜕𝑥 𝐢 + 𝜕𝑦 𝐣
𝜕𝑥 𝜕𝑦
 If f is a function of three variables x, y and z, then the gradient of f is the vector function
𝜕𝑓 𝜕𝑓 𝜕𝑓 𝜕𝑓 𝜕𝑓 𝜕𝑓
𝛻𝑓 defined by 𝛻𝑓 𝑥, 𝑦 = , , = 𝜕𝑥 𝐢 + 𝜕𝑦 𝐣 + 𝜕𝑧 𝐤
𝜕𝑥 𝜕𝑦 𝜕𝑧
 Notation: 𝛻𝑓 or grad𝑓
 Remarks: - Gradient represents the slope of the tangent of the graph of the function
- Magnitude of 𝛻𝑓 gives the slope of the graph of the function
Gradient vector
𝑥2 𝑦2
 Let 𝑓 𝑥, 𝑦 = + 4 . The gradient of f is
9
2 2
𝛻𝑓 𝑥, 𝑦 = 9 𝑥 𝐢 + 4𝑦 𝐣
 At point (0, 2), the gradient of f is

𝛻𝑓 = 0𝐢 + 𝐣 = 𝐣 = 0,1,0
The slope of the graph of f at (0, 2) is
02 + 12 + 02 = 1
The Fundamental Theorem of Line
Integrals
 Let C be a smooth curve given by the vector function 𝐫(𝑡), 0 ≤ 𝑡 ≤ 𝑏. Let f be a

differentiable function of two or three variables whose gradient vector 𝛻𝑓 is continuous
on C. Then ‫𝑏 𝐫 𝑓 = 𝐫𝑑 ∙ 𝑓𝛻 𝐶׬‬ − 𝑓(𝐫 𝑎 )
 Note: This theorem applies only for conservative vector fields
The Fundamental Theorem of Line
Integrals
 Let a curve C be defined by the parametric equations 𝑥 = 𝑡 2 + 1, 𝑦 = 𝑡 3 + 𝑡, 0 ≤ 𝑡 ≤ 1.

 Values of a function f is given in the table:
y 0 1 2
x
0 1 6 4
1 3 5 7
2 8 2 9
 Determine ‫𝐫𝑑 ∙ 𝑓𝛻 𝐶׬‬
 Solution: Evaluate x and y at the endpoints (that is when t = 0 and t = 1). Then, calculate
𝑓 𝐫 1 − 𝑓(𝐫 0 ). Answer: 6

Line Integral

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LINE INTEGRAL

Students should be able to

 The base of a circular fence is given by 𝑥 = cos(𝑡), 𝑦 = sin 𝑡 , 0 ≤ 𝑡 ≤ 2𝜋.

• Let say you are given a smooth curve C on R3

 Integrals along C in R2:

 Integrals along C in R3:

 Suppose now that C is a piecewise-smooth curve; that

 We can interpret the line integral of a positive

 If 𝑓(𝑥, 𝑦) ≥ 0, then ‫𝑥 𝑓 𝐶׬‬, 𝑦 𝑑𝑠 represents

 The base of a circular fence is given by

 f ( x, y ) dx  lim  f ( xk* , yk* )xk

 f ( x, y ) dy  lim  f ( xk* , yk* )yk

 Integrals along C in R3:

 f ( x, y, z ) dx  lim  f ( xk* , yk* , z k* )xk

 f ( x, y, z ) dy  lim  f ( xk* , yk* , z k* )yk

 f ( x, y, z ) dz  lim  f ( xk* , yk* , z k* )z k

 Adding the values of these integrals, we obtain 24.5 – 15 = 9.5

 At point (0, 2), the gradient of f is

 Let C be a smooth curve given by the vector function 𝐫(𝑡), 0 ≤ 𝑡 ≤ 𝑏. Let f be a

 Let a curve C be defined by the parametric equations 𝑥 = 𝑡 2 + 1, 𝑦 = 𝑡 3 + 𝑡, 0 ≤ 𝑡 ≤ 1.

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