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Double integral

  1. Apr 28, 2010 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    find [tex]\int\int_D sin\left(\frac{y}{x}\right)dA[/tex] bounded by [tex]x=0, y=\pi, x=y^2[/tex]


    3. The attempt at a solution

    I've only studied calculus 1, this problem is for my friend. I did read up briefly on double integrals however and this is why I'm stuck:

    From the limits and where the graphs intersect, we have:

    [tex]\int_0^{\pi^2}\int_{\sqrt{x}}^{\pi}sin\left(\frac{y}{x}\right)dydx[/tex]

    then integrating and evaluating the inside part:

    [tex]\int_0^{\pi^2}\left(-xcos\left(\frac{\pi}{x}\right)+xcos\left(\frac{1}{\sqrt{x}}\right)\right)dx[/tex]

    But finding the integral of that seems impossible. I also tried reversing the order of integration, but come up with the same problem.
     
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  3. Apr 28, 2010 #2

    gabbagabbahey

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    I think you'll have to make use of the special function

    [tex]\text{Ci}(z)\equiv-\int_z^\infty\frac{\cos(t)}{t}dt[/tex]

    called the Cosine integral


    Just split the integral into two and make the substitution [itex]u=\frac{\pi}{x}[/itex] for the first, and [itex]u=\frac{1}{\sqrt{x}}[/itex] for the second.
     
  4. Apr 28, 2010 #3

    Mentallic

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    Ok, let me try this...

    [tex]u=\frac{\pi}{x}, x=\frac{\pi}{u}, dx=\frac{-\pi}{u^2}du[/tex]

    [tex]v=\frac{1}{\sqrt{x}}, x=\frac{1}{v^2}, dx=\frac{-2}{v^3}dv[/tex]

    So substituting all this in:

    [tex]\pi^2\int_0^{\pi^2}\frac{cosu}{u^3}du+2\int_{\pi^2}^{0}\frac{cosv}{v^5}dv[/tex]

    But I'm unsure how to apply the fact that [tex]Ci(z)=\int\frac{cosz}{z}dz[/tex] to this expression.
     
  5. Apr 28, 2010 #4

    gabbagabbahey

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    First, you need to change your limits of integration, since ytou are no longer integrating over [itex]x[/itex]. Second, there's no need too use two different variables since [itex]\int_a^b f(v)dv=\int_a^b f(u)du[/itex] (i.e. for definite integral, the integration variable is essentially a dummy variable)

    You should end up with

    [tex]\int\int_D \sin\left(\frac{y}{x}\right)dA=-\pi^2\int_{\frac{1}{\pi}}^\infty\frac{\cos u}{u^3}du+2\int_{\frac{1}{\pi}}^\infty\frac{\cos u}{u^5}du[/tex]

    Use integration by parts twice on each.
     
  6. Apr 28, 2010 #5

    Mentallic

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    Oh yes of course, thanks that's really helpful :smile:

    Obviously integrating by parts will give a big long expression and looking at it now, I'm not even going to think about posting it here. Taking [itex]\infty[/itex] as one of the limits of the integration cancels out a lot.

    But one more thing, can [tex]cos\left(\frac{1}{\pi}\right)[/tex] be expressed more simply? And how do I evaluate the [tex]Ci\left(\frac{1}{\pi}\right)[/tex]?
     
  7. Apr 28, 2010 #6

    gabbagabbahey

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    No.

    Same way you evaluate [itex]e[/itex] (the base of the natural logarithm) or [itex]\pi[/itex]; approximate it numerically to arbitrary precision using a power series. In other words, just leave it as is. It can't be simplified in terms of elementary functions.
     
  8. Apr 28, 2010 #7

    Mentallic

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    Yes of course, but it's just that I don't know what the power series for Ci(z) is and I'm afraid my calculator doesn't have a button for it either :biggrin:

    And sometimes including numerical approximations are a nice addition to the big long expression.
     
  9. Apr 29, 2010 #8

    gabbagabbahey

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    Just use a better calculator:smile:

    I'm sure Wolfram alpha (Google it) will have no problem giving you a numerical approximation. Alternatively, I'd venture a guess that you could look up the power series online and find it quickly.
     
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