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Homework Help: Conservative and irrotational vectors

  1. Jan 25, 2010 #1
    1. Please can someone assist me in solving the attached vector problem.

    I have made several attempt to no avail


    Attached Files:

  2. jcsd
  3. Jan 25, 2010 #2


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    You've almost got part (i) finished. You just have to keep going. You got to this point:

    [tex]J=\int_C (ye^{xy}dx+(xe^{xy}+1)dy[/tex]

    Then you introduced a parameter u and apparently wrote x=u and y=u3 since C is the curve [itex]y=x^3[/itex]. This gave you dx=du and dy=3u2du, so you got to here:

    [tex]J=\int_{u_1}^{u_2} (ye^{u^4}du+3u^3e^{u^4}du+3u^2du)[/tex]

    So far, so good. You just need to substitute for y in terms of u in the first term and figure out what u1 and u2 are. You were told the contour runs from x=1 to x=6. Since you know how x and u are related, you can easily see what u1 and u2 equal.

    All the terms can easily be integrated either directly or with a simple substitution
  4. Jan 26, 2010 #3
    Thanks for your response.

    I still cannot solve the problem.

    Please let someone assist.
  5. Jan 26, 2010 #4


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    Where are you haveing trouble?
    You already had
    [tex]J=\int_{u_1}^{u_2} (ye^{u^4}du+3u^3e^{u^4}du+3u^2du)[/tex]
    and vela pointed out that, since [itex]y= u^3[/itex], that is the same as
    [tex]J=\int_{u_1}^{u_2} (u^3e^{u^4}du+3u^3e^{u^4}du+3u^2du)[/tex]
    so it is really just
    [tex]J=\int_{u_1}^{u_2} (4u^3e^{u^4}du+3u^2du)[/tex]

    You are told that x runs from 1 to 6. When x= 1 what is u? When x= 6, what is u?

    If you still can't do it, please show your work so we can see exactly where you have a problem.
  6. Jan 27, 2010 #5
    Hello everyone

    Please find attached the output of my integration and substitution.

    However, I am not convinced that this is correct.

    Please assist me.

    Attached Files:

  7. Jan 29, 2010 #6
    Please let someone assist me by reviewing my attached solution.

    I don't think I'm correct.

    Please I am counting on you.
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