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Double Integral

  1. Oct 15, 2009 #1
    Evaluate the integral by changing to polar coordinates

    [tex]\int\int arctan(y/x)[/tex]

    Given that

    0 [tex]\leq[/tex] x [tex]\leq[/tex] 1 and 0 [tex]\leq[/tex] y [tex]\leq[/tex] x

    Now I've changed the integral to

    [tex]\int\int \theta r dr d\theta[/tex]

    Such that 0 [tex]\leq[/tex] [tex]\theta[/tex] [tex]\leq[/tex] [tex]\frac{\pi}{2}[/tex] and 0 [tex]\leq[/tex] r [tex]\leq[/tex] [tex]\sqrt{2}[/tex]

    And evaluating this I get [tex]\frac{\pi^{2}}{8}[/tex]

    I don't think this is correct though, I have found out that the answer is [tex]\frac{\pi}{8}[/tex] - 0.25ln(2)

    Can someone show me where I'm going wrong please, thanks!

    Sorry for the poor Latex-ing I've never used it before.
     
  2. jcsd
  3. Oct 15, 2009 #2

    tiny-tim

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    Hi Juggler123! :smile:

    (have an integral: ∫ and a pi: π and a theta: θ and a ≤ :wink:)

    Sorry, but your limits are toooootally wrong. :redface:

    0 ≤ x ≤ 1 and 0 ≤ y ≤ x is a 45º triangle with base 1.

    So your limits are 0 ≤ θ ≤ 45º and 0 ≤ r ≤ … (depends on θ :wink:) ? :smile:
     
  4. Oct 15, 2009 #3
    Would it be correct to say that the upper limit for r is 1/cos[tex]\theta[/tex] ?
     
  5. Oct 15, 2009 #4

    tiny-tim

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    Hi Juggler123! :smile:

    (what happened to that θ i gave you? :confused:)
    Noooo. :redface:
     
  6. Oct 15, 2009 #5
    How can I go about finding the upper limit then? I used the fact that the maximum x=1=max.y then said x=rcos(theta) and rearranged, where have I gone wrong??
     
  7. Oct 15, 2009 #6

    Mark44

    Staff: Mentor

    Hint: In your triangular region, every x value on the right boundary has the same value.
     
  8. Oct 15, 2009 #7
    Sorry I'm still confused, why does knowing that on the right boundary x is always equal to 1?
     
  9. Oct 15, 2009 #8

    tiny-tim

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    oops!

    oh no, I had my diagram the wrong way up :redface:

    you were right, it is 0 ≤ r ≤ 1/cosθ.

    Sorry. :redface:

    ok, now integrate ∫∫ θr drdθ between those limits. :smile:
     
  10. Oct 15, 2009 #9

    Mark44

    Staff: Mentor

    Because, on that boundary x = 1 ==> rcos(θ) = 1, so r = 1/cos(θ).

    It looks like you already understand this, based on another reply you gave.
     
  11. Oct 15, 2009 #10
    Now I've stumbled across another problem!

    How would I integrate x/(2cos(x)^2) w.r.t x
     
  12. Oct 15, 2009 #11

    Mark44

    Staff: Mentor

    This is actually pretty simple.
    [tex]\int \frac{x}{2cos^2x} dx~=~1/2\int x sec^2 x dx[/tex]
    Edit: added missing x factor in right integral.
    Hint: There is a trig function whose derivative wrt x is sec^2(x).
     
    Last edited: Oct 15, 2009
  13. Oct 15, 2009 #12
    Isn't x/2cos(x)^2 equal to 0.5xsec(x)^2?? Not 0.5sec(x)^2
     
  14. Oct 15, 2009 #13

    Mark44

    Staff: Mentor

    Right you are. The integral should be
    [tex]\int \frac{x}{2cos^2x} dx~=~1/2\int x sec^2 x dx[/tex]
    This looks like a natural for integration by parts, with u = x and dv = sec^2(x)dx.
     
  15. Oct 15, 2009 #14
    Yes! That works brilliant, thanks for all the help!
     
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