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Math Help - 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

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

    As

    <br />
\begin{aligned}<br />
s(t)&=\int_a^t\,ds\\<br />
&=\int_a^t\,\sqrt{dx^2+dy^2+dz^2}\\<br />
&=\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}\\<br />
&=\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,<br />
\end{aligned}<br />

    we may conclude that

    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 C_1, C_2, and C_3 are as follows:

    <br />
\begin{array}{cccc}<br />
C_1: & x'=0 & y'=0 & z'=1 \\<br />
C_2: & x'=0 & y'=1 & z'=0 \\<br />
C_3: & x'=1 & y'=0 & z'=0<br />
\end{array}<br />

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

    \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 \textbf r(t) is different for each C:

    <br />
\begin{aligned}<br />
C_1:\,f(\textbf r(t))&=0+\sqrt{0}-t^2=-t^2 \\<br />
C_2:\,f(\textbf r(t))&=0+\sqrt{t}-1^2=\sqrt{t}-1 \\<br />
C_3:\,f(\textbf r(t))&=t+\sqrt{1}-1^2=t.<br />
\end{aligned}<br />
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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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