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Area inside r = 2 cos(θ) but outside r = 1

  1. Nov 14, 2009 #1
    1. The problem statement, all variables and given/known data
    Use double integrals to find the area inside the circle r = 2 cos(θ) and outside the circle r = 1.

    2. Relevant equations
    I figured this was too easy to require an graphic. If you can't picture the circles, imagine them in rectangular from:
    r = 2 cos(θ) ==> y2+(x-2)2=1
    r = 1 ==> y2+x2=1


    3. The attempt at a solution
    Both circles have a radius of 1 and you need to look at all 2\pi of the objects to see the full area of overlap. So this is what I tried:

    [tex]\int\stackrel{2\pi}{0}\int\stackrel{0}{1}[/tex] (2cos(θ)-r) drdθ

    The book says the answer is but I can't get it:

    [tex]\frac{\pi}{3}[/tex]+[tex]\frac{\sqrt{3}}{2}[/tex]
     
  2. jcsd
  3. Nov 14, 2009 #2
    The integral to find area in polar coordinates is:
    [tex]\iint_A r \, dr \, d\theta [/tex]

    Adjust the limits of integration to match the equations given. The actual contents of the integral ([tex]r \, dr \, d\theta[/tex]) will remain the same.
     
  4. Nov 14, 2009 #3

    LCKurtz

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    You need to find the polar coordinates of the two curves intersections and use appropriate limits. Not everything goes from 0 to [itex]2\pi[/itex] or 0 to 1.

    Generally to find an area using polar coordinate double integrals you need something like this:

    [tex]A = \int_{\theta_1}^{\theta_2} \int_{r_{inner}}^{r_{outer}}1\ rdrd\theta[/tex]

    and you need to determine the correct limits from your formulas and picture.
     
  5. Nov 14, 2009 #4
    Thanks, but that doesn't really get me any closer to an answer. Is the function I'm integrating correct? What are the ranges?
     
  6. Nov 14, 2009 #5

    LCKurtz

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    I can't make much sense out of your integrals. To get area with a double integral, you integrate the function 1. You need to look at the graphs. Find the [itex]\theta[/itex]'s where they intersect by setting the r values equal.
     
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