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) d: (a) ds = Σ dt (b) N = Σ If our vector field is F1 = 9xi 6yj, then we have the flux being (F1 N)ds. (c) F1(r(t)) Σ (d) As such, (F1 N) ds = Σ Now, if our vector field is F2 = 5xi + 3(x - y)j, then we have the flux being| (F2 N)ds.

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