Explanation Given: F=< y, z, x > , C is the triangle with vertices (0,0,0), (3,0,0) and (0,3,3) ; oriented counterclockwise. 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 ^ Evaluate: c u r l F → = ∇ → × F → = det ( i ^ j ^ k ^ ∂ ∂ x ∂ ∂ y ∂ ∂ z y z x ) =− i ^ − j ^ − k ^ Let us find the equation of the surface on which the curve C lies. The equation of a surface passing through three points (x 1, y 1 ,z 1), (x 2, y 2 ,z 2), (x 3, y 3 ,z 3) is given by: det ( x − x 1 y − y 1 z − z 1 x 2 − x 1 y 2 − y 1 z 2 − z 1 x 3 − x 1 y 3 − y 1 z 3 − z 1 ) = 0 = det ( x y z 3 0 0 0 3 3 ) = 0 ⇒ z = y Let D be the projection of the given surface S on the xy- plane then D is the circle the triangular region bounded by x=0, y=0 and x+y=3 :-

4th Edition

ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 18.2, Problem 15E

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

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Chapter 18.2, Problem 14E

Chapter 18.2, Problem 16E

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Chapter 18 Solutions

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∮ C F → · d R → using Stokes’ Theorem

Solution Summary: The author explains how Stokes' theorem is applied to the equation of the surface on which the curve C lies.

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To evaluate:

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

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Chapter 18 Solutions

∮ C F → · d R → using Stokes’ Theorem

Solution Summary: The author explains how Stokes' theorem is applied to the equation of the surface on which the curve C lies.

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

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Chapter 18 Solutions

To evaluate:

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