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Math Help - Line Integral along a straight line.

  1. #1
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    Line Integral along a straight line.

    I can do the problem, but my final number is wrong.

    If \(C\) is the line segment from \((6,1)\) to, \((0,0)\), <br />
find the value of the line integral:
    \(\int_C (8 y^2\,\vec i +  x\,\vec j)\cdot d\vec r =\)

    First we need to parametrize;
    \\r^{\rightarrow }(t)=(1-t)({p}_{1},{p}_{2})+t({q}_{1},{q}_{2})\\
    r^{\rightarrow }(t)=(1-t)(6,1)+t(0,0)\\
    r^{\rightarrow }(t)=(6-6t,1-t)+0t\\
    r^{\rightarrow }(t)=<6-6t,1-t>\\

    This means that the individual parametric equations are,

    x=6-6t,y=1-t

    The next step is to integrate;

    work=\int_{C}^{}{F}^{\rightarrow }*d{r}^{\rightarrow }=\int_{t=0}^{t=1}{F}^{\rightarrow }({r}^{\rightarrow }(t))*({r'}^{\rightarrow }(t))dt
    The problem:
    \int_{t=0}^{t=1}<8(1-t)^{2},(6-6t)>\bullet <-6,-1>dt
    \int_{t=0}^{t=1}48(1-t)^{2}+(-6+6t)dt
    \int_{t=0}^{t=1}(48(t)^{2}-96t+48-6+6t)dt
    \int_{t=0}^{t=1}(48(t)^{2}-102t+54)dt
    (16(t)^{3}-51t^{2}+54t)\mid_{0}^{1}

    So if you do that last line you should end up with a...(19),using the FTC 'b-a'. Of course in this case b=1,a=0.

    However, when I make it -19 I'm told I'm am correct; what have I done wrong to end up with the wrong sign.
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  2. #2
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    Quote Originally Posted by Zanderist View Post
    I can do the problem, but my final number is wrong.

    If \(C\) is the line segment from \((6,1)\) to, \((0,0)\), <br />
find the value of the line integral:
    \(\int_C (8 y^2\,\vec i +  x\,\vec j)\cdot d\vec r =\)

    First we need to parametrize;
    \\r^{\rightarrow }(t)=(1-t)({p}_{1},{p}_{2})+t({q}_{1},{q}_{2})\\
    r^{\rightarrow }(t)=(1-t)(6,1)+t(0,0)\\
    r^{\rightarrow }(t)=(6-6t,1-t)+0t\\
    r^{\rightarrow }(t)=<6-6t,1-t>\\

    This means that the individual parametric equations are,

    x=6-6t,y=1-t

    The next step is to integrate;

    work=\int_{C}^{}{F}^{\rightarrow }*d{r}^{\rightarrow }=\int_{t=0}^{t=1}{F}^{\rightarrow }({r}^{\rightarrow }(t))*({r'}^{\rightarrow }(t))dt
    The problem:
    \int_{t=0}^{t=1}<8(1-t)^{2},(6-6t)>\bullet <-6,-1>dt
    \int_{t=0}^{t=1}48(1-t)^{2}+(-6+6t)dt


    Here it must be -48 ...

    Tonio



    \int_{t=0}^{t=1}(48(t)^{2}-96t+48-6+6t)dt
    \int_{t=0}^{t=1}(48(t)^{2}-102t+54)dt
    (16(t)^{3}-51t^{2}+54t)\mid_{0}^{1}

    So if you do that last line you should end up with a...(19),using the FTC 'b-a'. Of course in this case b=1,a=0.

    However, when I make it -19 I'm told I'm am correct; what have I done wrong to end up with the wrong sign.
    .
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