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Green’s theorem
Use Green’s theorem to evaluate the line integral. Orient the curve counterclockwise unless
otherwise indicated
󰇛

󰇜


Where C is the rectangle with vertices (1,1), (3,1), (1,4) and (3,4).
Let C be positively be oriented, piecewise smooth, simple, closed curve and let D be the
region enclosed by the curve. If P and Q have continuous first order partial derivatives on D
then
 
󰇡




󰇢
Submitted by Albert Eshun
Solution
The graph below shows the area covered by C, where C is a rectangle with vertices (1,1),
(3,1), (1,4) and (3,4).
To solve this problem
Let
󰇛

󰇜
 
So,





According to Green’s theorem, the equation below is obtained after applying double integral
󰇡




󰇢
󰇛

󰇜

The integral is evaluated directly using the rectangular co-ordinates
󰇝
󰇛󰇜

󰇞
Which is given by ,
󰇛

󰇜

󰇛

󰇜

4
2
4
2
(1, 4)
(3, 4)
(1,1)
(3,1)
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󰇟
󰇠
󰇟
󰇛

󰇜
󰇛

󰇜

Therefore,
 
Reference
Zenisek, A.(1999). Green’s theorem from the viewpoint of applications, Application of
mathematics.44(1), 55-80.

Unformatted Attachment Preview

Green’s theorem Use Green’s theorem to evaluate the line integral. Orient the curve counterclockwise unless otherwise indicated ∫𝐶 (𝑙𝑛𝑥 + 𝑦)𝑑𝑥 − 𝑥 2 𝑑𝑦 Where C is the rectangle with vertices (1,1), (3,1), (1,4) and (3,4). Let C be positively be oriented, piecewise smooth, simple, closed curve and let D be the region enclosed by the curve. If P and Q have continuous first order partial derivatives on D then 𝜕𝑄 𝜕𝑃 ∬𝐶 𝑃𝑑𝑥 + 𝑄𝑑𝑦 = ∬𝐷 ( 𝜕𝑥 − 𝜕𝑦 ) 𝑑𝐴 Submitted by Albert Eshun Solution The graph below shows the area covered by C, where C is a rectangle with vertices (1,1), (3,1), (1,4) and (3,4). 4 (1, 4) (3, 4) 2 (3,1) (1,1) 2 4 To solve this problem Let 𝑃 = (𝑙𝑛𝑥 + 𝑦) 𝑎𝑛𝑑 𝑄 = −𝑥 2 So, 𝜕𝑄 𝜕𝑃 − 𝜕𝑦 = −2𝑥 − 1 𝜕𝑥 According to Green’s theorem, the equation below is obtained after applying double integral 𝜕𝑄 𝜕𝑃 I = ∬𝐷 (𝜕𝑥 − 𝜕𝑦 ) 𝑑𝐴 = ∬𝐷 (−2𝑥 − 1)𝑑𝐴 The integral is evaluated directly using the rectangular co-ordinates 𝐷 = {(𝑥, 𝑦)|1 ≤ 𝑥 ≤ 3, 1 ≤ 𝑦 ≤ 4} Which is given by , 3 4 3 𝐼 = ∫1 ∫1 (−2𝑥 − 1)𝑑𝑦𝑑𝑥 = ∫1 3(−2𝑥 − 1)𝑑𝑥 = 3[−𝑥 2 − 𝑥]13 = 3[(−9 − 3) − (−1 − 1) = −30 Therefore, ∮𝐶 𝐹. 𝑑𝑟 = −30 Reference Zenisek, A.(1999). Green’s theorem from the viewpoint of applications, Application of mathematics.44(1), 55-80. Name: Description: ...
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