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Transforming double integrals into Polar coordinates

  1. Jan 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that:

    [itex]I = \int\int_{T}\frac{1}{(1 + x^{2})(1 + y^{2})}dxdy = \int^{1}_{0}\frac{arctan(x)}{(1 + x^{2})}dx = \frac{\pi^{2}}{32}[/itex]

    where T is the triangle with successive vertices [itex](0,0), (1,0), (1,1)[/itex].

    *By transforming to polar coordinates [itex](r,θ)[/itex] show that:*

    [itex]I = \int^{\frac{\pi}{4}}_{0}\frac{log(\sqrt{2}cos(θ))}{cos(2θ)}dθ[/itex]


    2. Relevant equations



    3. The attempt at a solution
    Part one is fine:

    [itex]I = \int\int_{T}\frac{1}{(1 + x^{2})(1 + y^{2})}dxdy = \int^{1}_{0}\underbrace{\int^{x}_{0}\frac{1}{(1 + x^{2})(1 + y^{2})}dy}_{A}dx[/itex]

    [itex]A: \frac{1}{(1 + x^{2})}\int^{x}_{0}\frac{1}{(1 + y^{2})}dy = \frac{1}{(1 + x^{2})}\left[arctan(y)\right]^{x}_{0} = \frac{arctan(x)}{(1 + x^{2}}[/itex]

    [itex]\Rightarrow I = \int^{1}_{0}\frac{arctan(x)}{(1 + x^{2})}dx[/itex]

    [itex]Let:[/itex] [itex]u = arctan(x)[/itex]

    [itex]\frac{du}{dx} = \frac{1}{(1 + x^{2})} \Rightarrow du = \frac{1}{(1 + x^{2})}dx[/itex]

    [itex]x = 0 \Rightarrow u = 0[/itex]
    [itex]x = 1 \Rightarrow u = \frac{\pi}{4}[/itex]

    [itex]\Rightarrow I = \int^{\frac{\pi}{4}}_{0}udu = \left[\frac{1}{2}u^{2}\right]^{\frac{\pi}{4}}_{0} = \frac{\pi^{2}}{32}.[/itex]


    PART 2

    [itex]x = rcos(\vartheta)[/itex]
    [itex]y = rsin(\vartheta)[/itex]

    [itex]Boundaries:[/itex]
    [itex](i) y = 0 \Rightarrow rsin(θ) = 0[/itex]
    [itex](ii) x = 1 \Rightarrow rcos(θ) = 1[/itex]
    [itex](iii) y = x \Rightarrow rsin(θ) = rcos(θ) \Rightarrow tan(θ) = 1 \Rightarrow θ = \frac{\pi}{4}[/itex]

    From then on everything seems to get very complex and I'm not really sure what the limits are either.

    I was wondering if it was possible to go simply from:

    [itex]I = \int^{1}_{0}\frac{arctan(x)}{(1 + x^{2})}dx[/itex]

    to:

    [itex]I = \int^{\frac{\pi}{4}}_{0}\frac{log(\sqrt{2}cos(θ))}{cos(2θ}dθ[/itex]

    Please help :(
     
  2. jcsd
  3. Jan 24, 2013 #2

    Simon Bridge

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    OK - you want to show that transforming to polar coordinates does this:$$I = \int\int_{T}\frac{1}{(1 + x^{2})(1 + y^{2})}dxdy \rightarrow \int^{\frac{\pi}{4}}_{0}\frac{\log(\sqrt{2}\cos \theta)}{\cos 2 \theta }d\theta$$Is that supposed to be a natural logarithm on the RHS?

    You have noticed that:
    [itex]x = r\cos\theta[/itex]
    [itex]y = r\sin\theta[/itex]

    Boundaries:
    [itex]y = 0 \Rightarrow r\sin\theta = 0[/itex]
    [itex]x = 1 \Rightarrow r\cos\theta = 1[/itex]
    [itex]y = x \Rightarrow r\sin\theta = r\cos(\theta) \Rightarrow \tan\theta = 1 \Rightarrow \theta = \frac{\pi}{4}[/itex]

    You also need to realize that:
    ##dx.dy = dA = r.dr.d\theta## (check)

    Transforming the integrand would involve using the right trig identities - equip yourself with a table of them, and see that you are looking for something with ##\cos 2 \theta## ... the logarithm probably comes from integrating over ##r##.

    Please show your working.
     
  4. Jan 25, 2013 #3
    [itex]dx.dy=dA=r.dr.dθ[/itex] thats a given

    not sure about the limits on r...

    [itex](expanding gives:[/itex]

    [itex]\int^{\frac{\pi}{4}}_{0}\int^{?}_{0}\frac{r}{(1 + r^{2}sin^{2}(θ) + r^{2}cos^{2}(θ) + r^{4}sin^{2}(θ)cos^{2}(θ))}drdθ[/itex]

    [itex]= \int^{\frac{\pi}{4}}_{0}\int^{?}_{0}\frac{r}{(1 + r^{2} + r^{4}sin^{2}(θ)cos^{2}(θ))}drdθ[/itex]

    [itex]= \int^{\frac{\pi}{4}}_{0}\int^{?}_{0}\frac{1}{\frac{1}{r} + r + r^{3}sin^{2}(θ)cos^{2}(θ))}drdθ[/itex]

    then it gets confusing unless maybe you make a nice substitution?
     
  5. Jan 25, 2013 #4
    Try u = r^2

    Regarding the limits, r is the distance from (0, 0) to (x, y) on the the x = 1 line.
     
  6. Jan 25, 2013 #5
    When me and my friend first tried we got the upper limit as sqrt(2)?

    Unless its still a function in θ
     
  7. Jan 25, 2013 #6
    My friend and I*
     
  8. Jan 25, 2013 #7
    sqrt(2) can be the distance to ONE point; you have a segment from (1, 0) to (1, 1), so the distance must be a function of the polar angle.
     
  9. Jan 25, 2013 #8
    The boundary x = 1 gives:
    rcos(θ) = 1 which is:
    r = 1/cos(θ)

    Hmmm
     
    Last edited: Jan 25, 2013
  10. Jan 25, 2013 #9
    That seems OK.
     
  11. Jan 25, 2013 #10
    integral = -1/2 sec(2 theta) (log(r^2 (-cos(2 theta))+r^2+2)-log(r^2 cos(2 theta)+r^2+2))+constant
     
  12. Jan 25, 2013 #11
    You have demonstrated mastery of LATEX, so please use it for complex expressions.

    Please show how you obtained that result.
     
  13. Jan 25, 2013 #12
    LATEX is very difficult on a mobile sorry :(

    Thanks for the mastery compliment :)

    As for how I obtained the result... WolframAlpha

    But it times out when you try to put limits from 0 to 1/cos(θ)
     
  14. Jan 25, 2013 #13
    I am sure you can plug in the integration limits manually.
     
  15. Jan 25, 2013 #14
    It came out perfectly :D
     
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