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An inconsistent conservative vector field

  1. Jan 22, 2016 #1
    < y^2, 2xy+ e^(3z), 3ye^(3z)> is the vector field.

    the above vector field is inside an open simply connected domain.

    the parametric equations all have a continuous first order derivative inside the domain.

    Lastly, the curl of the vector field is <0, 0, 0>

    Thus, the vector field is conservative and is path independent.


    If i take the

    x axis, y axis, and z axis,
    Y axis, x axis, and z axis, OR
    Y axis, z axis, and x axis

    pathways, then I get the scalar field equation to be

    [f(x,y,z)] = xy^2 + y + ye^(3z)

    This is NOT CORRECT!!!!

    If I take the three axis pathways in any other order than the order listed above, then I get the correct answer.

    Correct function is
    [f(x,y,z)] = xy^2 + ye^(3z)

    There is not doubt that the vector field is conservative.

    If a vector field is conservative, then the path integral MUST BE path independent.

    Why do I get two different answers?...
  2. jcsd
  3. Jan 23, 2016 #2


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    Homework Helper

    Probably because you made a computation error. How can we tell without seeing the computation?
    Can you show how you got ##xy²+y+ye^{3z}## in one case and ##xy²+ye^{3z}## in an other case?
  4. Jan 23, 2016 #3
    X axis- y axis - z axis pathway.

    1. x axis

    Interval is from x=0 to x=x1

    Integrate <0, 1, 0> • <dx, 0, 0> to get 0

    2. Y-axis

    x=x1, dx,dz,z=0

    Interval is from y=0 to y=y1

    Integrate <y^2, 2x1y + 1, 3y> • <0, dy, 0> to get
    x1y^2 + y

    3. Z-axis
    x=x1, y=y1, dx=dy=0

    Interval is from z = 0 to z = z1
    Integrate <(y1)^2, 2x1y1 + e^(3z1), 3y1e^(3z)> • <0,0,dz> to get


    The total path integral is NOT CORRECT!!!

    [f(x1, y1, z1)] = Constant + x1y^2 + y + y1e^(3z1)

    Where did I go wrong?!?! Help
  5. Jan 23, 2016 #4
    Also... The same inconsistency occurs for this conservative vector field.

    <e^y , xe^y>

    Having e in the parametric equation seems to make the line integral path DEPENDENT.

  6. Jan 23, 2016 #5


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    In the last integral in the z-direction you miss a ##-y_1## term.
    ##\displaystyle \int_0^{z_1} 3y_1e^{3z} \, dz = \left. y_1e^{3z} \right|_0^{z_1} = y_1e^{3z_1}-y_1##
    Last edited: Jan 23, 2016
  7. Jan 23, 2016 #6
    Yes! Lol. Thanks
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