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Consider the semicircle r(t) = a cos(t)i + asin(1)j with 0 ≤ ≤ and a > 0. For a given vector field F, the flux across r(t) is is [(F.N) ds. Σ dt. (a) ds = a (b) N= If our vector field is F₁ = 7xi - 4yj, then we have the flux being (c) F₁ (r(t)) <7'a*cos(t), -4'a'sin(t)> (F₁-N) ds = (d) As such, Σ (e) F₂(r(t)) = <3'a'cos(t), 7(a*cos(t)-a*sin(t))> (1) As such, (F₂N) ds -2pia^2 Σ Now, if our vector field is F₂ = 3xi +7(x - y)j, then we have the flux being (F₂-N)ds. Σ [(F₁-N)ds. Σ W

Consider the semicircle r(t) = a cos(t)i + asin(1)j with 0 ≤ ≤ and a > 0. For a given vector field F, the flux across r(t) is is [(F.N) ds. Σ dt. (a) ds = a (b) N= If our vector field is F₁ = 7xi - 4yj, then we have the flux being (c) F₁ (r(t)) <7'a*cos(t), -4'a'sin(t)> (F₁-N) ds = (d) As such, Σ (e) F₂(r(t)) = <3'a'cos(t), 7(a*cos(t)-a*sin(t))> (1) As such, (F₂N) ds -2pia^2 Σ Now, if our vector field is F₂ = 3xi +7(x - y)j, then we have the flux being (F₂-N)ds. Σ [(F₁-N)ds. Σ W

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Consider the semicircle r(t) = a cos(t)i + a sin(1)j with 0 ≤ ≤ and a > 0. For a given vector field F, the flux across r(t) is is (a) ds = a (b) N= <cos(t), sin(t)> If our vector field is F₁ = 7xi-4yj, then we have the flux being (d) As such, Σ dt. (F₁-N) ds = (c) F₁ (r(t)) <7'a*cos(t), -4*a* sin(t)> · SCF₁-1 Now, if our vector field is F₂ = 3xi +7(x - y)j, then we have the flux being (F₂-N)ds. (e) F₂ (r(t)) = <3*a*cos(t), 7(a*cos(t)-a*sin(t))> Σ (1) As such, (F₂N) ds -2pia^2 Σ Σ [(F₁-N)ds. Σ W -N) ds.
Transcribed Image Text:Consider the semicircle r(t) = a cos(t)i + a sin(1)j with 0 ≤ ≤ and a > 0. For a given vector field F, the flux across r(t) is is (a) ds = a (b) N= <cos(t), sin(t)> If our vector field is F₁ = 7xi-4yj, then we have the flux being (d) As such, Σ dt. (F₁-N) ds = (c) F₁ (r(t)) <7'a*cos(t), -4*a* sin(t)> · SCF₁-1 Now, if our vector field is F₂ = 3xi +7(x - y)j, then we have the flux being (F₂-N)ds. (e) F₂ (r(t)) = <3*a*cos(t), 7(a*cos(t)-a*sin(t))> Σ (1) As such, (F₂N) ds -2pia^2 Σ Σ [(F₁-N)ds. Σ W -N) ds.
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