Explanation Given: F=<yz, xy, xz>, C is the square with vertices(0,0,2),(1,0,2),(1,1,2)and(0,1,2). Calculation: By Stokes’ theorem, ∮ C F → · d R → = ∬ S ( c u r l F → · N ^ ) d S where C is the boundary (orientation of C is compatible with that of S ) of the surface S whose upward unit normal vector is N ^ Let us now evaluate c u r l F → = ∇ → × F → = det ( i ^ j ^ k ^ ∂ ∂ x ∂ ∂ y ∂ ∂ z y z x y x z ) =( ∂ ∂ y ( x z ) − ∂ ∂ z ( x y ) ) i ^ − ( ∂ ∂ x ( x z ) − ∂ ∂ z ( y z ) ) j ^ + ( ∂ ∂ x ( x y ) − ∂ ∂ y ( y z ) ) k ^ = ( y − z ) i ^ − ( y − z ) k ^ C lies on the surface z=2 .Let D be the projection of the given surface S on the xy -plane then D is the square bounded by the lines x=0,x=1,y=0 and y=1 :-

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable

4th Edition

ISBN: 9781319312886

Author: Rogawski, Jon, Adams, Colin, FRANZOSA, Robert

Publisher: W. H. Freeman

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Chapter 17.2, Problem 12E

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$\oint_{C} \vec{F} \cdot d \vec{R}$ using Stokes’ Theorem

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Chapter 17.2, Problem 11E

Chapter 17.2, Problem 13E

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Chapter 17 Solutions

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable

Ch. 17.1 - Prob. 1PQ Ch. 17.1 - Prob. 2PQ Ch. 17.1 - Prob. 3PQ Ch. 17.1 - Prob. 4PQ Ch. 17.1 - Prob. 5PQ Ch. 17.1 - Prob. 1E Ch. 17.1 - Prob. 2E Ch. 17.1 - Prob. 3E Ch. 17.1 - Prob. 4E Ch. 17.1 - Prob. 5E

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∮ C F → · d R → using Stokes’ Theorem

Solution Summary: The author explains the Stokes’ theorem, where C is the square with vertices and a downward unit normal vector.

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To evaluate:

$\oint_{C} \vec{F} \cdot d \vec{R}$ using Stokes’ Theorem

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Chapter 17 Solutions

∮ C F → · d R → using Stokes’ Theorem

Solution Summary: The author explains the Stokes’ theorem, where C is the square with vertices and a downward unit normal vector.

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To evaluate: ∮CF→·dR→ using Stokes’ Theorem

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Chapter 17 Solutions

To evaluate:

$\oint_{C} \vec{F} \cdot d \vec{R}$ using Stokes’ Theorem