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Area of a polar Region

  1. May 14, 2006 #1

    G01

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    I have to find the area of the region enclosed by both of these curves.
    [tex] r=\sin\theta [/tex]

    [tex] r= \cos\theta [/tex]

    The area I need to find is the area of the overlap region of these to circles. The circles intersect at [tex] \pi /4[/tex]. Other than that I cant see what I have to do to set this problem up. Thanks for the help.
     
  2. jcsd
  3. May 14, 2006 #2
    One way to do it is to transform the problem into Cartesian coordinates....
     
  4. May 14, 2006 #3

    arildno

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    The area in question lies above the sine arc from [itex]0\leq\theta\leq\frac{\pi}{4}[/itex]
    and beneath the cosine arc, [itex]\frac{\pi}{4}\leq\theta\leq\frac{\pi}{2}[/itex]

    The simplest way to compute this area A is to compute as follows:
    [tex]A=\int_{0}^{\frac{\pi}{2}}\int_{0}^{\cos\theta}rdrd\theta-\int_{0}^{\frac{\pi}{4}}\int_{\sin\theta}^{\cos\theta}rdrd\theta[/tex]
     
  5. May 14, 2006 #4
    Yeah thats easier..I forgot :yuck:
     
  6. May 14, 2006 #5

    G01

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    I'm sorry aldridno, I'm not quite sure what you are doing there. I am just finishing my second semester of Calculus using Stewart's book, Chapter 11. Thats the background I have. Or maybe I know how to do it but I just dont understand what you are doing.
     
  7. May 14, 2006 #6
    Well what arildno's done is that he's used polar coordinates to represent an area element as [itex]r dr d\theta[/itex] rather than [itex]dxdy[/itex]. If you're uncomfortable with this, note that r in polar coordinates is always positive so [itex]r = \sin\theta[/itex] is the locus of a point where [itex]0<r\leq1[/itex] and [itex]0<\sin\theta \leq 1[/itex] (similar considerations apply for the other locus).

    Now you can convert the problem into cartesian coordinates by making sure you get the quadrants right.
     
  8. May 14, 2006 #7

    arildno

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    DRAW the region in question!!
    I just saw there was a similar thread by you on this, where ToxicBug gives you an alternative explanation.
    His is, indeed, correct, and yields the same answer as my method.
    To see this, we have the identity:
    [tex]\int_{0}^{\frac{\pi}{2}}\int^{\cos\theta}_{0}rdrd\theta=\int_{0}^{\frac{\pi}{4}}\int_{\sin\theta}^{\cos\theta}rdrd\theta+\int_{0}^{\frac{\pi}{4}}\int_{0}^{\sin\theta}rdrd\theta+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_{0}^{\cos\theta}rdrd\theta[/tex]
    Therefore:
    [tex]\int_{0}^{\frac{\pi}{2}}\int^{\cos\theta}_{0}rdrd\theta-\int_{0}^{\frac{\pi}{4}}\int_{\sin\theta}^{\cos\theta}rdrd\theta=\int_{0}^{\frac{\pi}{4}}\int_{0}^{\sin\theta}rdrd\theta+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_{0}^{\cos\theta}rdrd\theta[/tex]
    which is basically ToxicBug's answer.
     
    Last edited: May 14, 2006
  9. May 14, 2006 #8

    G01

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    Thanks, sorry about the double post, I've been sick and I guess I just didnt understand what Toxic Bug was saying. Thanks Alot.
     
  10. May 14, 2006 #9

    GCT

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    I think that we can use single integrals here. A double integral is employed for the volume, since one is summing up the value of z times the area element; although one can use the double integral to find the average area as well as other lower dimension variables.

    G01, integrate the area that is above the x axis, and subtract this from the region that's below the x axis. For the reasoning behind this method, review for yourself the fundamental theorem of calculus, above the x axis the area of integration is positive.

    So you can start by showing your work, if you still need assistance with this problem.
     
    Last edited: May 14, 2006
  11. May 14, 2006 #10

    GCT

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    You can actually integrate with respect to the individual curves. For instance from pi/4 to 3pi/2 sum positive valued areas while negating those portions that are below the x axis, you'll probably need to do this component wise. The same with the sine as well as cosine, and finally sum up all of the values, making sure to negate the negative portions.
     
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