Explanation Given: The condition is Green’s theorem is a special case of Stoke’s theorem Calculation: Green’s theorem: Let D be a domain whose boundary ∂ D is a simple closed curve, oriented counter clock wise. If P ( x , y ) and Q ( x , y ) are differentiable and have continuous first partial derivatives, then ∮ c P d x + Q d y = ∬ ( ∂ Q ∂ x − ∂ P ∂ y ) d A Or, In other words, for F = 〈 P , Q 〉 over a curve C ∫ C F . d s ∫ c P d x + Q d y = ∬ ( ∂ Q ∂ x − ∂ P ∂ y ) d A = ∬ D c u r l ( F ) d A Now, c u r l ( F ) = | i j k ∂ ∂ x ∂ ∂ y ∂ ∂ z P Q O | = − ∂ Q ∂ z i + ∂ P ∂ z j + ( ∂ Q ∂ x − ∂ P ∂ y ) k c u r l ( F ) ...

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Chapter 18.2, Problem 36E

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Chapter 18.2, Problem 35E

Chapter 18.2, Problem 37E

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Chapter 18 Solutions

CALCULUS (CLOTH)

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Why Green’s theorem is a special case of Stoke’s theorem

Solution Summary: The author explains that Green's theorem is a special case of Stoke’s theory.

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Chapter 18 Solutions