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Need some double integral help.

  1. Feb 13, 2007 #1

    double integral (e^(y^3)) dy dx

    Where dy is evaluated from sqrt(x/3) to 1

    .....and dx is evaluated from 0 to 3.

    I am lost.
    I don't even know how to start.:frown:
  2. jcsd
  3. Feb 13, 2007 #2


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    do u know how to do integrals?
    double integral is just normal integrals in disguise, so when ou are doing the "y"-integral, you treat all "x"s in the integrand as just a constant, and while doing the "x"-integral you treat all "y"s as constant. Then, everything would then be exactly the same as a normal single integral. In many cases, whether you do the x or y integral first does not matter.
    in your case i think you should do y first
  4. Feb 13, 2007 #3


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    The "y" integration doesn't look pretty. It's basically

    [tex] \int_{a}^{b} e^{y^3}{}dy [/tex]
  5. Feb 13, 2007 #4
    Hey guys yes I do know how to do integrals...however I don't know how to do this one. If I integrate y first I have to integrate e^(y^3)dy....and i don't know how to do this.

    If I integrate x first then I'll end up with having to
    integrate 3e^(y^3)dy....and I again I don't know how to do this.

  6. Feb 13, 2007 #5
    Here it is.

    [tex]\int_{0}^{3} \int_{sqrt(x/3)}^{1} e^{y^3} {}dy} {}dx [/tex]

    Yay!! I got the latex code....however this is the only thing I know about this problem.
    Last edited: Feb 13, 2007
  7. Feb 13, 2007 #6
    Nevermind....I got it. Thanks guys.
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