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Evaluate the line integral: int c (x+2y)dx +(x-y)dy where c is the curve x=2cost,...
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To answer this question, split up the line integral into two pieces:
intc (x + 2y)dx and intc (x - y)dy.
Our parameter is t, 0<=t<=pi/4 (I assume, because your problem statement gives inf <-- t < 0, which diverges )
We need to convert dx, dy into dt:
x = 2 cos t --> dx = -2 sin t dt
y = 4 sin t --> dy = 4 cos t
intc (x + 2y)dx = int [ (2 cos t + 8 sin t) ( -2 sin t ) dt , 0<= t <=pi/4]
= int [ -4 costsint - 16sint^2, 0<= t <=pi/4 ]
= intc (x - y)dy = int [ (2 cos t - 4 sin t) ( 4 cos t ) dt , 0<= t <=pi/4]
= int [ 8 cost^2 - 16 costsint, 0<= t <=pi/4 ]
intc (x + 2y)dx + intc (x - y)dy = int[ 8 cost^2 - 20 costsint - 16 sint^2, 0<= t <=pi/4 ]
=-4 t + 5 cos(2 t) + 6 sin(2 t) + C, evaluated from 0<=t<=pi/4
= 1 - pi = 2.14
Posted by kjcdb8er on January 24, 2011 at 5:15 AM (Answer #1)
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