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Line integral question

  1. Dec 3, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the line integral of ∫ x+yz dx + 2x dy + xyz dz

    C consists of line segments from (1,0,1) to (2,3,1) and from (2,3,1) to (2,5,2).


    2. Relevant equations

    r=(1-t)<r0> + t<r1> 0<t<1

    3. The attempt at a solution

    I split up the two line segments into C1 and C2.

    For C1 I got the parametric equation of x(t)=1+t, y(t)=3t, z(t)=1
    I plugged this into the original equation also using dx=dt, dy=3dt, dz=0
    This gave me the answer C1=13.

    I then found the parametric equations of C2 to be x(t)=2, y(t)=3-2t, z(t)=1-t.
    I plugged these into the equations again with their derivatives and got the answer to be -22/3.

    I added C1 and C2 and got 17/3 as my answer.

    The correct answer should be 97/3.
     
  2. jcsd
  3. Dec 4, 2011 #2
    On the first, if I let:

    x=1+t
    y=3t
    z=1

    I get 12.

    On the second, how about let

    x=2
    y=3+2t
    z=1+t
     
  4. Dec 4, 2011 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    If you had shown how you got this, we might have been able to point out a mistake. As it is, we, like jackmell, can only say that your first integral is wrong.

    Okay, when t= 0, (x, y, z)= (2, 3, 1) and when t= -1, (x, y, z)= (2, 5, 2). Did you then integrate from 0 to -1?

     
  5. Dec 4, 2011 #4
    Sorry I didn't explain my work. I integrated both C1 and C2 from 0 to 1.. why are you integrating it from -1 to 0?

    And here's the rest of my work,

    C1:
    r=(1-t)<1,0,1> + t<2,3,1>
    r=<1+t,3t,1>

    (from 0 to 1)∫ (1+4t) (1)dt +
    (from 0 to 1)∫ (2) (1+t) (3) dt +
    (from 0 to 1)∫ (1+t)(3t)(1)(0) dt

    [2t^2+t](0 to 1) +
    6[t+(t2/2)](0 to 1) +
    [t](0 to 1)

    (2 + 1) + (6 + 3) + 1 = 13


    C2:
    r=(1-t)<2,3,1> + t<2,5,2>
    r=<2,3-2t,1-t>

    (from 0 to 1)∫ 2+(3+2t)(1-t)(0)dt +
    (from 0 to 1)∫ 4(-2)dt +
    (from 0 to 1)∫ (2)(3-2t)(1-t)(-1) dt

    2[t](0 to 1) +
    -8[t](0 to 1) +
    -2[3t - (5/2)*t2 + (2/3)*t3](0 to 1)

    adding these up i got (-19/3) when I redid the problem again.

    so C1 + C2 = (20/3)
     
  6. Dec 5, 2011 #5
    Ok, that third one should be zero because z=1 which means dz=0 right? There you go. Got 36/3 of them.

    [/quote]

    Now that second one you have:

    [tex]\mathop\int\limits_{(2,3,2)}^{(2,5,2)} (x+yz)dx+2xdy+xyzdz[/tex]

    Ok, if I let:

    x=2
    y=3+2t
    z=1+t

    then I need to let t go from 0 to 1 so it's

    [tex]\int_0^1 (x+yz)dx+2xdy+xyzdz,\quad x(t)=2, y(t)=3+2t, z(t)=1+t[/tex]

    you can do that I bet.
     
  7. Dec 5, 2011 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I'm not. I asked if you did because your parametric equations, x(t)=2, y(t)=3-2t, z(t)=1-t, give x(0)= 2, y(0)= 3, z(0)= 1, the first point, and x(-1)= 2, y(-1)= 5, z(-1)= 2, the second point.

    If you integrate from 0 to 1 then, because x(1)= 2, y(1)= 3- 2= 1, z(1)= 1- 1= 0, you are integrating from (2, 3, 1) to (2, 1, 0) which is not what you wanted to do.
     
  8. Dec 5, 2011 #7


    Ok I think I understand..

    So what you're saying is...

    [itex]0,-1\int[/itex]
     
  9. Dec 5, 2011 #8
    Ok I think I understand..

    So what you're saying is...

    (from 0 to -1)∫ (2 + (3-2t)(1+2t))(0)dt +
    (from 0 to -1)∫ (2(2t))(3)dt +
    (from 0 to -1)∫ (2)(3-2t)(2)dt
    simplified:
    4 * (from 0 to -1)∫ -4t^2 + 4t + 1 dt

    4[(-4t^3)/3 + 2t^2 + t] dt <== from 0 to -1

    4(4/3 + 2 - 1 - (0 + 0 + 0)) = 28/3

    but,
    28/3 + 13 ≠ 97/3
     
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