Consider the semicircle r(t) = a cos(t)i + a sin(t)j with 0 0. For a given vector field F, the flux across r(t) is (F· N) ds. (a) ds = a Σ dt (b) N = Σ If our vector field is F1 = 10xi – 7yj, then we have the flux being (F1. N)ds. (c) F1 (r(t)) (10a(cost))i-(7a(sint))i Σ (d) As such, (F N) ds = ((10a(cost))i-(7a(sint)))(ac E Now, if our vector field is F, = 3zi + 9(* – ylj, then we have the flux being (F2 · N)ds. (e) F; (г(+)) — Σ (f) As such, (F2· N) ds = Σ

Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
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Consider the semicircle r(t)
= a cos(t)i + a sin(t)j with 0 <t < 7 and a > 0. For a given vector field F, the flux across r(t) is (F· N) ds.
(a) ds = a
Σ dt
(b) N =
<cost,sint>
Σ
If our vector field is F1 = 10xi – 7yj, then we have the flux being
(F1. N)ds.
(c) F1 (r(t)) (10a(cost))i-(7a(sint))i
Σ
(d) As such,
(F N) ds = ((10a(cost))i-(7a(sint)))(ac E
Now, if our vector field is F, = 3zi + 9(* – ylj, then we have the flux being
(F2 · N)ds.
(e) F; (г(+)) —
Σ
(f) As such,
(F2· N) ds =
Σ
Transcribed Image Text:Consider the semicircle r(t) = a cos(t)i + a sin(t)j with 0 <t < 7 and a > 0. For a given vector field F, the flux across r(t) is (F· N) ds. (a) ds = a Σ dt (b) N = <cost,sint> Σ If our vector field is F1 = 10xi – 7yj, then we have the flux being (F1. N)ds. (c) F1 (r(t)) (10a(cost))i-(7a(sint))i Σ (d) As such, (F N) ds = ((10a(cost))i-(7a(sint)))(ac E Now, if our vector field is F, = 3zi + 9(* – ylj, then we have the flux being (F2 · N)ds. (e) F; (г(+)) — Σ (f) As such, (F2· N) ds = Σ
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