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The flux of the vector field  along the outward normal, across the  circle x2 + y2 = 1 is equal to,
  • a)
    0
  • b)
  • c)
    π
  • d)
    3π/4
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The flux of the vector fieldalong the outward normal, across thecircle...
The parametrization  Traces the circle counterclock wise exactly once. We can therefore use this parametrization in the eqn,
With M = x - y = cost - sint 
N = x = cost 
and dy = d(sint) = cost dt 
dx = d(cost) = -sint dt 
we find
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Most Upvoted Answer
The flux of the vector fieldalong the outward normal, across thecircle...
The parametrization  Traces the circle counterclock wise exactly once. We can therefore use this parametrization in the eqn,
With M = x - y = cost - sint 
N = x = cost 
and dy = d(sint) = cost dt 
dx = d(cost) = -sint dt 
we find
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