Use Green's Theorem to evaluate the line integral along the given positively oriented curve.
1 answer:
Answer:
∫ C ( y + e√x) dx + ( 2x + cosy² ) dy = 1/3
Step-by-step explanation: See Annex
Green Theorem establishes:
∫C ( Mdx + Ndy ) = ∫∫R ( δN/dx - δM/dy ) dA
Then
∫ C ( y + e√x) dx + ( 2x + cosy² ) dy
Here
M = 2x + cosy² δM/dy = 1
N = y + e√x δN/dx = 2
δN/dx - δM/dy = 2 - 1 = 1
∫∫(R) dxdy ∫∫ dxdy
Now integration limits ( see Annex)
dy is from x = y² then y = √x to y = x² and for dx
dx is from 0 to 1 then
∫ dy = y | √x ; x² ∫dy = x² - √x
And
∫₀¹ ( x² - √x ) dx = x³/3 - 2/3 √x |₀¹ = 1/3 - 0
∫ C ( y + e√x) dx + ( 2x + cosy² ) dy = 1/3
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Step-by-step explanation:
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