1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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!).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted