ENGINEERING MATHEMATICS II
LINE INTEGRALS
,1 • Line Integral of a Scalar Field
• Line Integral of a Vector
2 Field
• Line Integral
3 Independent of
LINE INTEGRALS
LEARNING UNIT 3
, LINE INTEGRAL OF A SCALAR FIELD
• To begin, we find a way to integrate a function (a scalar field) along a path.
• Let x :[a ,b] two points on a curve 𝒞. Let f(x, y) a continuous function
• And 𝑑𝑠 a infinitely small changes of the arc length on the curve 𝒞
𝑑𝑠 = 𝑑𝑥 2 + 𝑑𝑦 2
𝑑𝑡
Multiplying by
𝑑𝑡
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑥 2 𝑑𝑦 2
𝑑𝑠 = 𝑑𝑠 = 𝑑𝑥 2 + 𝑑𝑦 2 ⟹ 𝑑𝑠 = 𝑑𝑠 = + 𝑑𝑡
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 2 𝑑𝑡 2
Finally
𝑑𝑥 2 𝑑𝑦 2
𝑑𝑠 = + 𝑑𝑡 𝑜𝑟 𝑑𝑠 = 𝑥′ 𝑡 2 + 𝑦 ′ 𝑡 2 𝑑𝑡
𝑑𝑡 2 𝑑𝑡 2
Then
𝑡=𝑏
න 𝑓 𝑥, 𝑦 𝑑𝑠 = න 𝑓 𝑥 𝑡 ,𝑦 𝑡 𝑥′ 𝑡 2 + 𝑦 ′ 𝑡 2 𝑑𝑡
𝒞 𝑡=𝑎
, LINE INTEGRAL OF A SCALAR FIELD
Let 𝑓(𝑥, 𝑦, 𝑧) a continuous function
𝑑𝑠 = 𝑑𝑥 2 + 𝑑𝑦 2 + 𝑑𝑧 2
𝑑𝑡
Multiplying by
𝑑𝑡
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑥 2 𝑑𝑦 2 𝑑𝑧 2
𝑑𝑠 = 𝑑𝑠 = 𝑑𝑥 2 + 𝑑𝑦 2 + 𝑑𝑧 2 ⟹ 𝑑𝑠 = 𝑑𝑠 = + + 𝑑𝑡
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 2 𝑑𝑡 2 𝑑𝑡 2
Finally
𝑑𝑥 2 𝑑𝑦 2 𝑑𝑧 2
𝑑𝑠 = + + 𝑑𝑡 𝑜𝑟 𝑑𝑠 = 𝑥′ 𝑡 2 + 𝑦′ 𝑡 2 + 𝑧 ′ 𝑡 2 𝑑𝑡
𝑑𝑡 2 𝑑𝑡 2 𝑑𝑡 2
Then
𝑡=𝑏
න 𝑓 𝑥, 𝑦, 𝑧 𝑑𝑠 = න 𝑓 𝑥 𝑡 , 𝑦 𝑡 , 𝑧(𝑡) 𝑥′ 𝑡 2 + 𝑦′ 𝑡 2 + 𝑧 ′ 𝑡 2 𝑑𝑡
𝒞 𝑡=𝑎
LINE INTEGRALS
,1 • Line Integral of a Scalar Field
• Line Integral of a Vector
2 Field
• Line Integral
3 Independent of
LINE INTEGRALS
LEARNING UNIT 3
, LINE INTEGRAL OF A SCALAR FIELD
• To begin, we find a way to integrate a function (a scalar field) along a path.
• Let x :[a ,b] two points on a curve 𝒞. Let f(x, y) a continuous function
• And 𝑑𝑠 a infinitely small changes of the arc length on the curve 𝒞
𝑑𝑠 = 𝑑𝑥 2 + 𝑑𝑦 2
𝑑𝑡
Multiplying by
𝑑𝑡
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑥 2 𝑑𝑦 2
𝑑𝑠 = 𝑑𝑠 = 𝑑𝑥 2 + 𝑑𝑦 2 ⟹ 𝑑𝑠 = 𝑑𝑠 = + 𝑑𝑡
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 2 𝑑𝑡 2
Finally
𝑑𝑥 2 𝑑𝑦 2
𝑑𝑠 = + 𝑑𝑡 𝑜𝑟 𝑑𝑠 = 𝑥′ 𝑡 2 + 𝑦 ′ 𝑡 2 𝑑𝑡
𝑑𝑡 2 𝑑𝑡 2
Then
𝑡=𝑏
න 𝑓 𝑥, 𝑦 𝑑𝑠 = න 𝑓 𝑥 𝑡 ,𝑦 𝑡 𝑥′ 𝑡 2 + 𝑦 ′ 𝑡 2 𝑑𝑡
𝒞 𝑡=𝑎
, LINE INTEGRAL OF A SCALAR FIELD
Let 𝑓(𝑥, 𝑦, 𝑧) a continuous function
𝑑𝑠 = 𝑑𝑥 2 + 𝑑𝑦 2 + 𝑑𝑧 2
𝑑𝑡
Multiplying by
𝑑𝑡
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑥 2 𝑑𝑦 2 𝑑𝑧 2
𝑑𝑠 = 𝑑𝑠 = 𝑑𝑥 2 + 𝑑𝑦 2 + 𝑑𝑧 2 ⟹ 𝑑𝑠 = 𝑑𝑠 = + + 𝑑𝑡
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 2 𝑑𝑡 2 𝑑𝑡 2
Finally
𝑑𝑥 2 𝑑𝑦 2 𝑑𝑧 2
𝑑𝑠 = + + 𝑑𝑡 𝑜𝑟 𝑑𝑠 = 𝑥′ 𝑡 2 + 𝑦′ 𝑡 2 + 𝑧 ′ 𝑡 2 𝑑𝑡
𝑑𝑡 2 𝑑𝑡 2 𝑑𝑡 2
Then
𝑡=𝑏
න 𝑓 𝑥, 𝑦, 𝑧 𝑑𝑠 = න 𝑓 𝑥 𝑡 , 𝑦 𝑡 , 𝑧(𝑡) 𝑥′ 𝑡 2 + 𝑦′ 𝑡 2 + 𝑧 ′ 𝑡 2 𝑑𝑡
𝒞 𝑡=𝑎