4 The flux of the curl of the vector field F(x, y, z) = (xy + e¯³², xz − ze³¹², xyz), (x, y, z) € R³ through the surface S = {(x, y, z) Є R³ : (x+3)² + y² + z² = 27, x ≥ 0} oriented such that its normal vector has a positive second component, equals: (A) -27π (B) 0 (C) 18π (D) 27π
4 The flux of the curl of the vector field F(x, y, z) = (xy + e¯³², xz − ze³¹², xyz), (x, y, z) € R³ through the surface S = {(x, y, z) Є R³ : (x+3)² + y² + z² = 27, x ≥ 0} oriented such that its normal vector has a positive second component, equals: (A) -27π (B) 0 (C) 18π (D) 27π
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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The correct answer is C,
i know that we need to use stokes theorem and parametrize the equations then write the equation F with respect to the curve but i cant seem to find a way to do it, the integral should be from 0 to 2pi but i might be wrong
could you show me the steps to get to 18pi
![4 The flux of the curl of the vector field F(x, y, z) = (xy + e¯³², xz − ze³¹², xyz), (x, y, z) €
R³ through the surface S = {(x, y, z) Є R³ : (x+3)² + y² + z² = 27, x ≥ 0} oriented such
that its normal vector has a positive second component, equals:
(A) -27π
(B) 0
(C) 18π
(D) 27π](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6099d21a-e15a-47f8-adbb-0c871c33581f%2F66491b36-d101-43bf-94d8-ae90556f35cd%2Fwor85n7_processed.png&w=3840&q=75)
Transcribed Image Text:4 The flux of the curl of the vector field F(x, y, z) = (xy + e¯³², xz − ze³¹², xyz), (x, y, z) €
R³ through the surface S = {(x, y, z) Є R³ : (x+3)² + y² + z² = 27, x ≥ 0} oriented such
that its normal vector has a positive second component, equals:
(A) -27π
(B) 0
(C) 18π
(D) 27π
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