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Math Help - Path Integral problems

  1. #1
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    Path Integral problems

    Have a problem that i have had a go at but am completely unsure if am anywhere near on the right track any help would be appreciated.


    Evaluate the path integral;

    <br />
\int_{C} f . ds<br /> <br />

    Where f = xy;

    And the path C is the part of the circle centred at the origin starting at (0,2) going anticlockwise ending at (-2,0).

    I did that x=2cost, y=2sint;
    so that c(t) = (2cost , 2sint),
    where 0< t < pi/2

    then did;

    <br /> <br />
\int f.ds = \int c(t).[c'(t)].dt<br /> <br />

    but am not sure i have set this up properly am a bit lost here.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by monster View Post
    Have a problem that i have had a go at but am completely unsure if am anywhere near on the right track any help would be appreciated.


    Evaluate the path integral;

    <br />
\int_{C} f . ds<br /> <br />

    Where f = xy;

    And the path C is the part of the circle centred at the origin starting at (0,2) going counterclockwise ending at (-2,0).

    I did that x=2cost, y=2sint;
    so that c(t) = (2cost , 2sint),
    where 0< t < pi/2

    then did;

    <br /> <br />
\int f.ds = \int c(t).[c'(t)].dt<br /> <br />

    but am not sure i have set this up properly am a bit lost here.
    Well, one thing that's a bit unclear is that you have f(x,y)=xy, yet you are only parameterizing two variables.

    I think you want c(t)=(2\cos(t),2\sin(t),0), \frac{\pi}{2}\leq t\leq \pi
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  3. #3
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    Actually, I think the first problem is in the limits for t. As you go from (0,2) to (-2,0), that is, with c(t)=(2\cos{t},2\sin{t}), then \frac{\pi}{2}\le{t}\le\pi.
    Next x(t)=2\cos{t}, and y(t)=2\sin{t}, so f(c(t))=xy=4\sin{t}\cos{t}.
    And since this is the line integral of a scalar function, ds is the arc length parameter:
    ds^2=dx^2+dy^2, so in terms of t, ds=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{  dy}{dt}\right)^2}\,dt=\sqrt{(-2\sin{t})^2+(2\cos{t})^2}\,dt
    ds=\sqrt{4\sin^2{t}+4\cos^2{t}}\,dt=2\,dt
    and you have
    \int_{\pi/2}^{\pi}(4\sin{t}\cos{t})2dt=8\int_{\pi/2}^{\pi}\sin{t}\cos{t}\,dt.

    --Kevin C.
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by redsoxfan325 View Post
    Well, one thing that's a bit unclear is that you have f(x,y)=xy, yet you are only parameterizing two variables.

    I think you want c(t)=(2\cos(t),2\sin(t),0), \frac{\pi}{2}\leq t\leq \pi
    ds = \sqrt{[dx/dt]^2+[dy/dt]^2+[dz/dt]^2}\,dt = 2\,dt

    8\int_{\pi/2}^{\pi}\sin(t)\cos(t)\,dt = 4\int_{\pi/2}^{\pi}\sin(2t)\,dt = -2\cos(2t)|_{\pi/2}^{\pi} = -2-2 = \boxed{-4}
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  5. #5
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    Where are you getting the z term, redsoxfan325? This is a two-dimensional problem (in the xy plane), so only x and y are involved; there's no z.

    --Kevin C.
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  6. #6
    Super Member redsoxfan325's Avatar
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    He/she said f=xy, but it doesn't change the answer either way because z=0.
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  7. #7
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    Quote Originally Posted by TwistedOne151 View Post
    Actually, I think the first problem is in the limits for t. As you go from (0,2) to (-2,0), that is, with c(t)=(2\cos{t},2\sin{t}), then \frac{\pi}{2}\le{t}\le\pi.
    Next x(t)=2\cos{t}, and y(t)=2\sin{t}, so f(c(t))=xy=4\sin{t}\cos{t}.
    And since this is the line integral of a scalar function, ds is the arc length parameter:
    ds^2=dx^2+dy^2, so in terms of t, ds=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{  dy}{dt}\right)^2}\,dt=\sqrt{(-2\sin{t})^2+(2\cos{t})^2}\,dt
    ds=\sqrt{4\sin^2{t}+4\cos^2{t}}\,dt=2\,dt
    and you have
    \int_{\pi/2}^{\pi}(4\sin{t}\cos{t})2dt=8\int_{\pi/2}^{\pi}\sin{t}\cos{t}\,dt.

    --Kevin C.
    I went;

    c(t) = (2cos(t) , 2sin(t))
    so c'(t) = ( - 2sin(t), 2cos(t))

    so [c'(t)] = 2
    so;
    <br /> <br />
\int_C f.ds = \int_{\pi/2}^{\pi} (8cos(t)sin(t)).dt<br />
    then made sub u = cost to give
     <br /> <br /> <br />
\int_{0}^{-1} -8u .du<br />

    = -4

    Is the way i have gone about it correct?
    and thank you for all help guys.
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  8. #8
    Super Member redsoxfan325's Avatar
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    Yeah, you did it correctly.
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