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Thread: Line Integrals

  1. #1
    s7b
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    Line Integrals

    I don't understand how to go about solving a problem of this form;

    Integrate f(x,y,z)=x+ sqrt(y) - z^2 over the path C from (0,0,0) to (1,1,1) given by C=C1nC2uC3, where

    C1: r(t)=tk , 0<t<1
    C2: r(t)=tj + k , 0<t<1
    C3: r(t)=ti + j + k , 0<t<1

    Anyone have any suggestions??
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  2. #2
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    If I understand correctly, the line integral for scalar functions is defined as

    $\displaystyle \int_C\,f(\textbf r(t))\,ds=\int_a^b\,f(\textbf r(t))s'(t)\,dt.$

    As

    $\displaystyle
    \begin{aligned}
    s(t)&=\int_a^t\,ds\\
    &=\int_a^t\,\sqrt{dx^2+dy^2+dz^2}\\
    &=\int_a^t\,\sqrt{\left(\frac{dx}{dt}dt\right)^2+\ left(\frac{dy}{dt}dt\right)^2+\left(\frac{dz}{dt}d t\right)^2}\\
    &=\int_a^t\,\sqrt{\left(\frac{dx}{dt}\right)^2+\le ft(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\righ t)^2}\,dt,
    \end{aligned}
    $

    we may conclude that

    $\displaystyle s'(t)=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\fr ac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2}.$

    Now, these derivatives on $\displaystyle C_1$, $\displaystyle C_2$, and $\displaystyle C_3$ are as follows:

    $\displaystyle
    \begin{array}{cccc}
    C_1: & x'=0 & y'=0 & z'=1 \\
    C_2: & x'=0 & y'=1 & z'=0 \\
    C_3: & x'=1 & y'=0 & z'=0
    \end{array}
    $

    In all cases, therefore, we end up with $\displaystyle s'(t)=1$. Our task is now to find

    $\displaystyle \int_{C_1}\,f(\textbf r(t))\,dt+\int_{C_2}\,f(\textbf r(t))\,dt+\int_{C_3}\,f(\textbf r(t))\,dt.$

    The function $\displaystyle \textbf r(t)$ is different for each $\displaystyle C$:

    $\displaystyle
    \begin{aligned}
    C_1:\,f(\textbf r(t))&=0+\sqrt{0}-t^2=-t^2 \\
    C_2:\,f(\textbf r(t))&=0+\sqrt{t}-1^2=\sqrt{t}-1 \\
    C_3:\,f(\textbf r(t))&=t+\sqrt{1}-1^2=t.
    \end{aligned}
    $
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  3. #3
    s7b
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    Ok, this makes sense.

    Is there a different approach when faced with a question like this;

    Integrate f over the given curve: f(x,y) = (x+y^2)/(sqrt(1+x^2)) , C: y=(x^2)/2 from (1,1/2) to (0,0)

    ..Or is the same method applied.
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