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Math Help - Line Integrals

  1. #1
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    Line Integrals

    If F = (x+4)i +xyj + (xz-y)k, evaluate \int c F.dr along the following paths C:

    the straight line joining (0,0,0) to (2,1,1)

    I don't really understand what to do and what the limits are.

    Thanks Adam
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by adam_leeds View Post
    the straight line joining (0,0,0) to (2,1,1)

    \gamma (t)=(2t,t,t)\quad (t\in [0,1])
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  3. #3
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    You can write your integral in a bit more standard notation this way:

    \int_{C}\mathbf{F}\cdot d\mathbf{r}. The code to produce this, for your convenience, was

    [TEX]\int_{C}\mathbf{F}\cdot d\mathbf{r}.[/TEX]
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  4. #4
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    Quote Originally Posted by FernandoRevilla View Post
    \gamma (t)=(2t,t,t)\quad (t\in [0,1])
    Thanks i've done this and got the right answer.

    What happens for the same question, if it is the straight lines from (0,0,0) to (0,0,1) then to (0,1,1) and then to (2,1,1)
    Last edited by adam_leeds; May 7th 2011 at 02:16 AM.
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Then,

    \displaystyle\int_C \vec{F}\cdot d\vec{r}=\displaystyle\int_{\gamma_1} \vec{F}\cdot d\vec{r}+\displaystyle\int_{\gamma_2} \vec{F}\cdot d\vec{r}

    where

    \gamma_1,\;\gamma_2

    are the corresponding lines.
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  6. #6
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    Quote Originally Posted by FernandoRevilla View Post
    Then,

    \displaystyle\int_C \vec{F}\cdot d\vec{r}=\displaystyle\int_{\gamma_1} \vec{F}\cdot d\vec{r}+\displaystyle\int_{\gamma_2} \vec{F}\cdot d\vec{r}

    where

    \gamma_1,\;\gamma_2

    are the corresponding lines.
    Is gamma 1 (0,0,0) to (0,0,1)

    And gamma 2 (0,0,1) and then to (2,1,1) ?
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  7. #7
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by adam_leeds View Post
    Is gamma 1 (0,0,0) to (0,0,1)

    And gamma 2 (0,0,1) and then to (2,1,1) ?

    Yes, those are the lines.
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  8. #8
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    Quote Originally Posted by FernandoRevilla View Post
    Yes, those are the lines.
    Ok

    Im not getting the same answer as my book

    (1-t)(0,0,0) + t(0,0,1) = (0,0,t)

    and (1-t)(0,1,1) + t(2,1,1) = (2t,1,1)

    So for the first one
    r' = (0,0,1)
    F.r' = 0

    So the integral of that is 0

    Then for the second one

    r' = (2,0,0)

    F.r' = 4t + 8

    so the integral between 1 and 0 = 10

    But the answer in my book is 38/3
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  9. #9
    MHF Contributor FernandoRevilla's Avatar
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    The second line is

    \gamma_2(t)=(2t,t,1)\quad (t\in [0,1])
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  10. #10
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    Quote Originally Posted by FernandoRevilla View Post
    The second line is

    \gamma_2(t)=(2t,t,1)\quad (t\in [0,1])
    How?

    (1-t)(0,1,1) + t(2,1,1) =(0+2t, 1-t+t, 1-t+t) = (2t,1,1)
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  11. #11
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by adam_leeds View Post
    How?

    (1-t)(0,1,1) + t(2,1,1) =(0+2t, 1-t+t, 1-t+t) = (2t,1,1)

    Look to your answers #4 and #6. The initial point of the second line is (0,0,1) and now, you have changed it.
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  12. #12
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    Quote Originally Posted by FernandoRevilla View Post
    Look to your answers #4 and #6. The initial point of the second line is (0,0,1) and now, you have changed it.
    Really sorry, i made a mistake in 4, it should be (0,1,1)
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