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Changing order of integration (double) in polar coord

  1. Sep 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Change the order of the limits of integration of the following double integral and evaluate.

    2. Relevant equations

    [itex] \int_{0}^\frac{\pi}{2} \int_{0}^{cos(\theta)} cos(\theta)\,dr\,d\theta [/itex]

    3. The attempt at a solution

    Evaluating as it is, I arrive an answer of [itex] \frac{\pi}{4} [/itex].

    I know the region to be integrated is the semicircle bounded by the polar axis, with corner points at r = 0, and r = 1, with a height of 1/2. I know that normally, in the cartesian case, to change the order of integration requires the limits to be written from y(x) to x(y), with the x or y intervals adjust accordingly.

    Thus, for this problem, the original region is bounded by:

    [itex] 0<r<cos(\theta)[/itex] and [itex] 0<\theta<\frac{\pi}{2} [/itex].

    Changing the form, I would write,

    [itex] 0<\theta<cos^{-1}(r) [/itex] and [itex] 0<r<1[/itex]

    Trying to evaluate in this manner, I end up at

    [itex] \int_{0}^1 sin(cos^{-1}(r))\,dr [/itex],

    after which I cannot go further. I have very little experience of changing limit orders in the polar case. Any hints would be appreciated!
     
    Last edited: Sep 5, 2013
  2. jcsd
  3. Sep 5, 2013 #2

    CompuChip

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    Hint: note that sin x = sqrt ( 1 - cos^2 x ).
     
  4. Sep 5, 2013 #3
    Cool, thanks; Using the hint: I get

    ## \int_{0}^1 \sqrt{1-cos(r)}\,dr ##

    Multiplying by a factor of sin(r)/sin(r), I get,

    ## \int_{0}^1 \frac{sin(r)}{\sqrt{1+cos(r)}}\,dr ##.

    I use a substitution of

    ## u = 1 + cos(r) ## with
    ## \,du = -sin(r)\,dr ##.

    I substitute the limits by using

    ## u_0 = 1+cos(0) = 2 ##
    ## u_1 = 1+cos(1) = ~1.5403... ##

    Evaluating the new integral, through

    ## -[2u^{\frac{1}{2}}] ##

    between u1=1.5403... and u0 = 2, I get an answer of ~0.173144 which is not quite ##\frac{\pi}{4} ##

    Are my limits wrong, or did I blunder somewhere else?
     
  5. Sep 6, 2013 #4

    CompuChip

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    Unfortunately the first line is still incorrect.

    What would x be, looking at your integrand (it's not just r, as you seem to have assumed!).
     
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