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Trouble with line integral

  1. May 7, 2012 #1
    Let $${\tmmathbf\vec{F} = yx^2 \tmmathbf\hat{i} + \sin (\pi y) \tmmathbf\hat{j}}$$, and let $${C}$$ be the curve along the line segment starting at (0,2) and ending at (1,4).

    $${\int_C \tmmathbf\vec{F} \cdot d \tmmathbf\vec{r} =}$$

    My path comes out to be:

    r=ti +2tj
    dr=i+2j;
    F=2t^3i+sin(2*pi*t)j
    Now F.dr=2(t^3)+2(sin(2*pi*t))
    so
    ∫2(t^3)+2(sin(2*pi*t))dt=(2t^4 - cos(2*pi*t)/pi ) where t goes from 0 to 1

    Note: r,dr and F are vectors.

    The answer from the above computation comes out to be 1/2.

    I've checked the answer and it's wrong so I've made a mistake somewhere. Could someone identify it?
     
  2. jcsd
  3. May 8, 2012 #2
    Let me rewrite the confusing part of my previous post:

    Let F=yx^2 i + sin(pi*y) j;

    and find

    ∫( on curve C) F.dr

    Now:

    My path comes out to be:

    r=ti +2tj
    dr=i+2j;
    F=2t^3i+sin(2*pi*t)j
    Now F.dr=2(t^3)+2(sin(2*pi*t))
    so
    ∫2(t^3)+2(sin(2*pi*t))dt=(2t^4 - cos(2*pi*t)/pi ) where t goes from 0 to 1

    Note: r,dr and F are vectors.

    The answer from the above computation comes out to be 1/2.

    I've checked the answer and it's wrong so I've made a mistake somewhere. Could someone identify it?
     
  4. May 8, 2012 #3

    LCKurtz

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    r = ti + 2tj doesn't pass through (0,2).
     
  5. May 9, 2012 #4
    If I'm not mistaken

    ∫2(t^3)+2(sin(2*pi*t))dt=(1/2)(t^4)-(cos(2*pi*t)/pi)

    you have =2t^4; by simply taking the derivative of this you would get 8t^3, not 2t^3.



    How does r(t)=ti+2tj not pass through (0,2)? To find the vector of a line segment, it's just <x2-x1, y2-y1>, is it not?
     
  6. May 9, 2012 #5

    HallsofIvy

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    Yes, but the vector is not the parametric equations.

    The parametric equations of a line with direction vector <a, b> , passing through [itex](x_0, y_0)[/itex] are [itex]x= at+ x_0[/itex] and [itex]y= bt+ y_0[/itex].

    The line given by it+ 2tj or x= t, y= 2t, passes through (0, 0) and (1, 2), parallel to the line in this problem.
     
  7. May 9, 2012 #6


    Some points for you to consider:

    1) Learn urgently to write with LaTeX in this forum

    2) The path you wrote is not a straight line between (0,2) and (1,4)...in fact, it doesn't even pass through neither of these

    two points! You must know how to produce the equation of a straight line between two points in the plane.

    3) After the above is done, and if I didn't make any error, you must get the integral [tex]\int_0^1\left(2t^3+2t^2+2\sin (\pi(2t+2))\right)dt=\frac{7}{6}[/tex]

    DonAntonio
     
  8. May 9, 2012 #7
    Thanks guys.
     
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