Let C represent a parabolic arc from the point (3, -3, 2) to (4, 1, 5) and
F(x, y, z) = [x/(x^2 + y^2 + z^2)] i + [y/(x^2 + y^2 + z^2)] j + [z/(x^2 + y^2 + z^2)] k.
Given this information, evaluate
∫ F •dr
c
A) (1/2) ln (19/7)
B) (1/2) ln (19/11)
C) (1/2) ln (19/12)
D) (1/2) ln (19/15)
E) (1/2) ln (35/22)
F) (1/2) ln (21/11)
G) (1/2) ln (17/13)
H) (1/2) ln (11/2)
I) none of these
F(x, y, z) = [x/(x^2 + y^2 + z^2)] i + [y/(x^2 + y^2 + z^2)] j + [z/(x^2 + y^2 + z^2)] k.
Given this information, evaluate
∫ F •dr
c
A) (1/2) ln (19/7)
B) (1/2) ln (19/11)
C) (1/2) ln (19/12)
D) (1/2) ln (19/15)
E) (1/2) ln (35/22)
F) (1/2) ln (21/11)
G) (1/2) ln (17/13)
H) (1/2) ln (11/2)
I) none of these
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Note that F(x,y,z) = ∇[(1/2) ln(x^2 + y^2 + z^2)].
Thus, the integral equals (by the fundamental theorem of calculus for line integrals)
(1/2) ln(x^2 + y^2 + z^2) {for (x,y,z) = (3, -3, 2) to (4, 1, 5)}
= (1/2) (ln 42 - ln 22)
= (1/2) ln(21/11).
I hope this helps!
Thus, the integral equals (by the fundamental theorem of calculus for line integrals)
(1/2) ln(x^2 + y^2 + z^2) {for (x,y,z) = (3, -3, 2) to (4, 1, 5)}
= (1/2) (ln 42 - ln 22)
= (1/2) ln(21/11).
I hope this helps!