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Homework Help: Double Integral with Strange Limits of Integration

  1. May 10, 2012 #1
    1. The problem statement, all variables and given/known data
    For the double integral ∫[0,1]∫[0,x^3] e^(y/x) dxdy
    (a) sketch the region of integration
    (b) evaluate the integral and
    (c) re-express the integral with the order of integration reversed

    2. Relevant equations

    3. The attempt at a solution
    The problem is that I've never seen a double integral problem with the limits of integration with respect to x in terms of a function of x, not y. I couldn't find any examples online or in my book where this is the case. The problem is written such that your boundaries should be 0≤x≤x^3 and 0≤y≤1... but x=x^3 doesn't make any sense to me as an upper boundary for x.

    I know how to do most double integral problems, reverse the order of integration, etc. but this has me stumped. Am I right in thinking this might be a typo? or is there some way to make sense of the region that I'm just not seeing?
  2. jcsd
  3. May 10, 2012 #2


    User Avatar
    Gold Member

    Hi bossman27 and welcome to PF!

    Using LaTeX, here is how your double integral appears:
    [tex]\int^1_0 \int^{x^3}_0 e^{(y/x)}\,.dxdy[/tex]
    But it should have been:
    [tex]\int^{x=1}_{x=0} \int^{y=x^3}_{y=0} e^{(y/x)}\,.dydx[/tex]
  4. May 10, 2012 #3


    Staff: Mentor

    0 <= x <= x3 doesn't make sense to me either. That should be 0 <= y <= x3, and 0 <=x <= 1.

    Are you sure you don't have dx and dy switched?
    This would make more sense.
    $$\int_{x = 0}^1\int_{y = 0}^{x^3}e^{y/x}dy~dx$$
  5. May 10, 2012 #4
    I'm sure I did not switch them, that was exactly how the problem appears on this practice test... I was thinking that it was probably a typo as well, but as the final is tomorrow morning I wanted to make sure I wasn't missing something.

    Thanks to both of you for the quick responses!
  6. May 10, 2012 #5
    It's entirely possible that you are essentially being asked for the anti-derivative, so you should think of the bound for the integral as a parameter, and the internal integral as being taken with respect to some dummy variable [itex]x^\prime[/itex].
  7. May 10, 2012 #6


    Staff: Mentor

    That's an interesting idea, but I don't think it's applicable here. bossman27 said that the integral was exactly as it appeared on the practice test. I believe that the instructor switched the order of dx and dy by mistake.
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