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Math Help - Having problems with another line integral

  1. #1
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    Having problems with another line integral

    int[(x+2y)dx + (x-2y)dy]; C is the line segment from (1,1) to (3,-1). The answer is 0. I got an answer of -8 by just taking the integral with respect to x before dx and respect to y after that. Then f(3,-1) - f(1,1). Obviously I did something wrong, any ideas?
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  2. #2
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    Quote Originally Posted by pakman View Post
    int[(x+2y)dx + (x-2y)dy]; C is the line segment from (1,1) to (3,-1). The answer is 0. I got an answer of -8 by just taking the integral with respect to x before dx and respect to y after that. Then f(3,-1) - f(1,1). Obviously I did something wrong, any ideas?
    I am not sure what you did, but you tought you can use the fundamental theorem of line integral you cannot.
    Because the cross partials test fails.

    The parametrization of the curve is,
    x=t and y=2-t for 1<=t<=3.

    That means,
    INT(1,3)(x+2y)*(dx/dt)dt+INT(1,3)(x-2y)(dy/dt)dt
    Insert the parametrization.

    INT(1,3)(t+2(2-t))*(dt/dt)dt+INT(1,3)(t-2(2-t))*(dy/dt)dt

    INT(1,3) 4-t dt + INT(1,3) 3t-4 dt

    What ever that is should be the answer.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    INT(1,3)(t+2(2-t))*(dt/dt)dt+INT(1,3)(t-2(2-t))*(dy/dt)dt
    INT(1,3) 4-t dt + INT(1,3) 3t-4 dt
    There is a minor sign error there: (dy/dt)=(-1).
    Now the answer is 0.
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