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Math Help - Line integral

  1. #1
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    Line integral

    To calculate the integral of ydx+(x^2+y^2)dy over the curve y=\sqrt{4-x^2} between (-2,0) and (2,0), can I parameterize the curve to x=2\cos t, y=2\sin t, \pi \leq t \leq 2\pi?
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  2. #2
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    Re: Line integral

    yep
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  3. #3
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    Re: Line integral

    Quote Originally Posted by polarbear73 View Post
    To calculate the integral of ydx+(x^2+y^2)dy over the curve y=\sqrt{4-x^2} between (-2,0) and (2,0), can I parameterize the curve to x=2\cos t, y=2\sin t, \pi \leq t \leq 2\pi?
    Note that the contour is an semicircle in the first and second quadrants in a clockwise direction.
    Therefore $y= 0\to 2\to 0~\&~x= -2\to 0\to 2$

    So your parametric path will not work. But $x=2\cos(t),~y=-2\sin(t),~\pi\le t\le 2\pi$ will work.
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  4. #4
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    Re: Line integral

    You're right but I bet dollars to donuts he means going through the 3rd and 4th quadrant and just left off a minus sign.
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  5. #5
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    Re: Line integral

    Quote Originally Posted by romsek View Post
    You're right but I bet dollars to donuts he means going through the 3rd and 4th quadrant and just left off a minus sign.
    I taught complex integrals so long that I recognize a standard trick question. The $y=\sqrt{4-x^2}$ is the upper half-circle in I & II.
    So $0\le y\le 2~\&~-2\le x\le 2$.
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  6. #6
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    Re: Line integral

    Hey thanks. I did think that the curve went through the third and fourth quadrants because I thought the standard convention of angular displacement was counterclockwise. I failed to notice that the curve, as written, had to be in the first and second quadrant.
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  7. #7
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    Re: Line integral

    Plato keeps us all honest... or tries anyway.
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  8. #8
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    Re: Line integral

    So if I integrate  ydx+(x^2+y^2)dy over the curve y=\sqrt{4-x^2} and parameterize the curve as x=2\cos t, y=2\sin t, 0\leq t \leq\pi I get \int(-4\sin^2 t +8\cos t) dt unless I'm doing something wrong. And if I do that integration I get -2t+8\sin t+\sin 2t+ C. If I evaluate that from 0 to  \pi I get -2\pi. This is not the answer in the text book. Could someone point me in the right direction?
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  9. #9
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    Re: Line integral

    Quote Originally Posted by polarbear73 View Post
    So if I integrate  ydx+(x^2+y^2)dy over the curve y=\sqrt{4-x^2} and parameterize the curve as x=2\cos t, y=2\sin t, 0\leq t \leq\pi I get \int(-4\sin^2 t +8\cos t) dt unless I'm doing something wrong. And if I do that integration I get -2t+8\sin t+\sin 2t+ C. If I evaluate that from 0 to  \pi I get -2\pi. This is not the answer in the text book. Could someone point me in the right direction?
    That not at all correct. Read reply #3.
    In the above you have gone from (2,0) to (-2,0) clockwise.
    That is not at all what the OP is asking for.
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