Double Integral

  • Thread starter Juggler123
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  • #1
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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.
 

Answers and Replies

  • #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:
 
  • #3
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Would it be correct to say that the upper limit for r is 1/cos[tex]\theta[/tex] ?
 
  • #4
tiny-tim
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Hi Juggler123! :smile:

(what happened to that θ i gave you? :confused:)
Would it be correct to say that the upper limit for r is 1/cos[tex]\theta[/tex] ?

Noooo. :redface:
 
  • #5
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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??
 
  • #6
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Hint: In your triangular region, every x value on the right boundary has the same value.
 
  • #7
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Sorry I'm still confused, why does knowing that on the right boundary x is always equal to 1?
 
  • #8
tiny-tim
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oops!

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??

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:
 
  • #9
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Sorry I'm still confused, why does knowing that on the right boundary x is always equal to 1?
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.
 
  • #10
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Now I've stumbled across another problem!

How would I integrate x/(2cos(x)^2) w.r.t x
 
  • #11
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Now I've stumbled across another problem!

How would I integrate x/(2cos(x)^2) w.r.t x
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:
  • #12
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Isn't x/2cos(x)^2 equal to 0.5xsec(x)^2?? Not 0.5sec(x)^2
 
  • #13
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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.
 
  • #14
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Yes! That works brilliant, thanks for all the help!
 

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