(c) Let D = R² \ {(0,0)} and consider the vector field defined on D defined as: X -Y x² + y² x² + y² (i) Show that g is conservative on D. g(x, y) = i+ $.9 C (ii) Let C be the unit circle around the origin. Suppose a fellow student proposes the following statement "Since C is closed and g is conservative, we must have: 9.ds = 0 j since integrals of conservative vector fields around closed loops are always zero.' 29 By calculating g.ds directly, show that your fellow student's state- ment is not true and explain why they were incorrect in their reasoning. [9 (iii) Let a and ß be two arbitrary positive real numbers and let y be an arbitrary smooth curve starting at (–a, 0), ending at (3,0) that only traverses the lower half of the xy-plane. Evaluate the integral: ds
(c) Let D = R² \ {(0,0)} and consider the vector field defined on D defined as: X -Y x² + y² x² + y² (i) Show that g is conservative on D. g(x, y) = i+ $.9 C (ii) Let C be the unit circle around the origin. Suppose a fellow student proposes the following statement "Since C is closed and g is conservative, we must have: 9.ds = 0 j since integrals of conservative vector fields around closed loops are always zero.' 29 By calculating g.ds directly, show that your fellow student's state- ment is not true and explain why they were incorrect in their reasoning. [9 (iii) Let a and ß be two arbitrary positive real numbers and let y be an arbitrary smooth curve starting at (–a, 0), ending at (3,0) that only traverses the lower half of the xy-plane. Evaluate the integral: ds
Algebra and Trigonometry (MindTap Course List)
4th Edition
ISBN:9781305071742
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter9: Vectors In Two And Three Dimensions
Section9.FOM: Focus On Modeling: Vectors Fields
Problem 11P
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i have seen that you have done part i of the question. i would like to ask if you can do part ii) and part iii) as well, since i dont see any solutions for those questions. i will add the image for the question again.
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