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Math Help - Line integrals and a Force vector

  1. #1
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    Line integrals and a Force vector

    Hey everyone,
    I have to evaluate F=<2+x2,3xy> where a curve is formed by moving along x2 + y2 = 4 from (2,0) to (0,2)
    then along the line segment from (0,2) to (-1,0)

    So far I've parameterized the first curve as r(t)=<2cost(t), 2sin(t)>, taking the derivative of that to get my tangent vector.
    Then I parameterized F(r(t)) to get F(r(t))=<2+ 4cos(t)^2, 12cos(t)sin(t)>
    Dotted my new F and tangent vector to get 16cos(t)2sin(t) - 4sin(t) and took the integral from 0 to pi/2
    For this I got: -28/3, wolfram got 4/3

    For the second curve I set r(t) = < -t, 2-2t> from 0 to 1 and got the tangent vector as <-1,-2>
    My new F is F(r(t))=<2+t^2, 6t2-6t>
    I dotted those to get 12t - 13t^2 - 2 and took the integral from 0 to 1
    For this I got -1/3 and wolfram got -1/3

    Adding these I got -29/3 and wolfram got 1, neither of these is what it's supposed to be, I guess the answer is -17/3 but I can't find how to get that anyway!
    All help is greatly appreciated! Really really confused...
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  2. #2
    Senior Member DeMath's Avatar
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    Re: Line integrals and a Force vector

    The first integral is easily calculated without the parameterization

    \begin{gathered}y = \sqrt {4 - {x^2}} ~~ \Rightarrow ~~dy = \frac{{ - x\,dx}}{{\sqrt {4 - {x^2}} }} \hfill \\[5pt] I_1= \int\limits_{L_1}(2 + x^2)\,dx + 3xy\,dy= \int\limits_2^0\!\left( {2 + {x^2} + 3x\sqrt {4 - {x^2}} \frac{{ - x}}{{\sqrt {4 - {x^2}} }}} \right)\!dx = 2\int\limits_2^0 (1-x^2)\,dx=  \ldots  = \frac{4}{3} \hfill \\ \end{gathered}

    To calculate the integral along the second line, first find the equation of the line (straight line)

    \begin{gathered}\frac{x-0}{-1-0} = \frac{y-2}{0-2}~~ \Leftrightarrow~~ y = 2x + 2~~ \Rightarrow~~ dy = 2\,dx \hfill \\ I_2= \int\limits_{L_2}(2 + {x^2})\,dx + 3xy\,dy = \int\limits_0^{ - 1} {(2 + x^2 + 3x(2x + 2)2)\,dx} }  = \int\limits_0^{ - 1}(2 + 12x + 13x^2)\,dx=  \ldots  =  - \frac{1}{3} \hfill \\\end{gathered}

    So, I=I_1+I_2=\frac{4}{3}+\left(-\frac{1}{3}\right)=1.
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  3. #3
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    Re: Line integrals and a Force vector

    Ah! I see now.
    thanks DeMath!
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