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: 9781319323394

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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Let F Use Stokes' Theorem to evaluate F. dr, where C is the triangle with vertices (6,0,0), (0,6,0), and (0,0,6), oriented counterclockwise as viewed from above.

= Use Stokes' Theorem to find the circulation of F 5y + 5zj+2ak around the triangle obtained by tracing out the path (6,0,0) to (6,0, 6), to (6, 3, 6) back to (6,0,0). Circulation = - dr = -45

4. Consider the vector function r(z, y) (r, y, r2 +2y"). (a) Re-write this vector function as surface function in the form f(1,y). (b) Describe and draw the shape of the surface function using contour lines and algebraic analysis as needed. Explain the contour shapes in all three orthogonal directions and explain and label all intercepts as needed. (c) Consider the contour of the surface function on the plane z= for this contour in vector form. 0. Write the general equation

Chapter 18 Solutions

CALCULUS W/SAPLING ACCESS >IC<

Ch. 18.1 - Prob. 1PQ Ch. 18.1 - Prob. 2PQ Ch. 18.1 - Prob. 3PQ Ch. 18.1 - Prob. 4PQ Ch. 18.1 - Prob. 5PQ Ch. 18.1 - Prob. 1E Ch. 18.1 - Prob. 2E Ch. 18.1 - Prob. 3E Ch. 18.1 - Prob. 4E Ch. 18.1 - Prob. 5E

Ch. 18.1 - Prob. 6E Ch. 18.1 - Prob. 7E Ch. 18.1 - Prob. 8E Ch. 18.1 - Prob. 9E Ch. 18.1 - Prob. 10E Ch. 18.1 - Prob. 11E Ch. 18.1 - Prob. 12E Ch. 18.1 - Prob. 13E Ch. 18.1 - Prob. 14E Ch. 18.1 - Prob. 15E Ch. 18.1 - Prob. 16E Ch. 18.1 - Prob. 17E Ch. 18.1 - Prob. 18E Ch. 18.1 - Prob. 19E Ch. 18.1 - Prob. 20E Ch. 18.1 - Prob. 21E Ch. 18.1 - Prob. 22E Ch. 18.1 - Prob. 23E Ch. 18.1 - Prob. 24E Ch. 18.1 - Prob. 25E Ch. 18.1 - Prob. 26E Ch. 18.1 - Prob. 27E Ch. 18.1 - Prob. 28E Ch. 18.1 - Prob. 29E Ch. 18.1 - Prob. 30E Ch. 18.1 - Prob. 31E Ch. 18.1 - Prob. 32E Ch. 18.1 - Prob. 33E Ch. 18.1 - Prob. 34E Ch. 18.1 - Prob. 35E Ch. 18.1 - Prob. 36E Ch. 18.1 - Prob. 37E Ch. 18.1 - Prob. 38E Ch. 18.1 - Prob. 39E Ch. 18.1 - Prob. 40E Ch. 18.1 - Prob. 41E Ch. 18.1 - Prob. 42E Ch. 18.1 - Prob. 43E Ch. 18.1 - Prob. 44E Ch. 18.1 - Prob. 45E Ch. 18.1 - Prob. 46E Ch. 18.1 - Prob. 47E Ch. 18.1 - Prob. 48E Ch. 18.1 - Prob. 49E Ch. 18.1 - Prob. 50E Ch. 18.1 - Prob. 51E Ch. 18.2 - Prob. 1PQ Ch. 18.2 - Prob. 2PQ Ch. 18.2 - Prob. 3PQ Ch. 18.2 - Prob. 4PQ Ch. 18.2 - Prob. 5PQ Ch. 18.2 - Prob. 1E Ch. 18.2 - Prob. 2E Ch. 18.2 - Prob. 3E Ch. 18.2 - Prob. 4E Ch. 18.2 - Prob. 5E Ch. 18.2 - Prob. 6E Ch. 18.2 - Prob. 7E Ch. 18.2 - Prob. 8E Ch. 18.2 - Prob. 9E Ch. 18.2 - Prob. 10E Ch. 18.2 - Prob. 11E Ch. 18.2 - Prob. 12E Ch. 18.2 - Prob. 13E Ch. 18.2 - Prob. 14E Ch. 18.2 - Prob. 15E Ch. 18.2 - Prob. 16E Ch. 18.2 - Prob. 17E Ch. 18.2 - Prob. 18E Ch. 18.2 - Prob. 19E Ch. 18.2 - Prob. 20E Ch. 18.2 - Prob. 21E Ch. 18.2 - Prob. 22E Ch. 18.2 - Prob. 23E Ch. 18.2 - Prob. 24E Ch. 18.2 - Prob. 25E Ch. 18.2 - Prob. 26E Ch. 18.2 - Prob. 27E Ch. 18.2 - Prob. 28E Ch. 18.2 - Prob. 29E Ch. 18.2 - Prob. 30E Ch. 18.2 - Prob. 31E Ch. 18.2 - Prob. 32E Ch. 18.2 - Prob. 33E Ch. 18.2 - Prob. 34E Ch. 18.2 - Prob. 35E Ch. 18.2 - Prob. 36E Ch. 18.2 - Prob. 37E Ch. 18.2 - Prob. 38E Ch. 18.3 - Prob. 1PQ Ch. 18.3 - Prob. 2PQ Ch. 18.3 - Prob. 3PQ Ch. 18.3 - Prob. 4PQ Ch. 18.3 - Prob. 5PQ Ch. 18.3 - Prob. 1E Ch. 18.3 - Prob. 2E Ch. 18.3 - Prob. 3E Ch. 18.3 - Prob. 4E Ch. 18.3 - Prob. 5E Ch. 18.3 - Prob. 6E Ch. 18.3 - Prob. 7E Ch. 18.3 - Prob. 8E Ch. 18.3 - Prob. 9E Ch. 18.3 - Prob. 10E Ch. 18.3 - Prob. 11E Ch. 18.3 - Prob. 12E Ch. 18.3 - Prob. 13E Ch. 18.3 - Prob. 14E Ch. 18.3 - Prob. 15E Ch. 18.3 - Prob. 16E Ch. 18.3 - Prob. 17E Ch. 18.3 - Prob. 18E Ch. 18.3 - Prob. 19E Ch. 18.3 - Prob. 20E Ch. 18.3 - Prob. 21E Ch. 18.3 - Prob. 22E Ch. 18.3 - Prob. 23E Ch. 18.3 - Prob. 24E Ch. 18.3 - Prob. 25E Ch. 18.3 - Prob. 26E Ch. 18.3 - Prob. 27E Ch. 18.3 - Prob. 28E Ch. 18.3 - Prob. 29E Ch. 18.3 - Prob. 30E Ch. 18.3 - Prob. 31E Ch. 18.3 - Prob. 32E Ch. 18.3 - Prob. 33E Ch. 18.3 - Prob. 34E Ch. 18.3 - Prob. 35E Ch. 18.3 - Prob. 36E Ch. 18.3 - Prob. 37E Ch. 18.3 - Prob. 38E Ch. 18.3 - Prob. 39E Ch. 18.3 - Prob. 40E Ch. 18.3 - Prob. 41E Ch. 18.3 - Prob. 42E Ch. 18.3 - Prob. 43E Ch. 18.3 - Prob. 44E Ch. 18 - Prob. 1CRE Ch. 18 - Prob. 2CRE Ch. 18 - Prob. 3CRE Ch. 18 - Prob. 4CRE Ch. 18 - Prob. 5CRE Ch. 18 - Prob. 6CRE Ch. 18 - Prob. 7CRE Ch. 18 - Prob. 8CRE Ch. 18 - Prob. 9CRE Ch. 18 - Prob. 10CRE Ch. 18 - Prob. 11CRE Ch. 18 - Prob. 12CRE Ch. 18 - Prob. 13CRE Ch. 18 - Prob. 14CRE Ch. 18 - Prob. 15CRE Ch. 18 - Prob. 16CRE Ch. 18 - Prob. 17CRE Ch. 18 - Prob. 18CRE Ch. 18 - Prob. 19CRE Ch. 18 - Prob. 20CRE Ch. 18 - Prob. 21CRE Ch. 18 - Prob. 22CRE Ch. 18 - Prob. 23CRE Ch. 18 - Prob. 24CRE Ch. 18 - Prob. 25CRE Ch. 18 - Prob. 26CRE Ch. 18 - Prob. 27CRE Ch. 18 - Prob. 28CRE Ch. 18 - Prob. 29CRE Ch. 18 - Prob. 30CRE Ch. 18 - Prob. 31CRE Ch. 18 - Prob. 32CRE Ch. 18 - Prob. 33CRE Ch. 18 - Prob. 34CRE Ch. 18 - Prob. 35CRE Ch. 18 - Prob. 36CRE Ch. 18 - Prob. 37CRE Ch. 18 - Prob. 38CRE Ch. 18 - Prob. 39CRE Ch. 18 - Prob. 40CRE Ch. 18 - Prob. 41CRE

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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.

Concept explainers

Videos

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