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Find Areas in Polar Coordinates

  1. May 14, 2010 #1
    I can't seem to get the correct answer. I rechecked my calculations but no luck. Any help is appreciated. Thanks.

    1. The problem statement, all variables and given/known data
    Find the area inside the larger loop and outside the smaller loop of the limacon below.
    r = sqrt(3)/2 + cos(theta)


    2. Relevant equations
    A = (integral from a to b) (1/2)*r^2 d@

    Half Angle Formula:
    (cos@)^2 = (1/2)(1+cos2@)

    3. The attempt at a solution
    A=2[ (integral from 0 to 2pi/3) [ (1/2)*(sqrt(3)/2 + cos@)^2 ]d@ - (integral from 2pi/3 to pi) [ (1/2)*(sqrt(3)/2 + cos@)^2 ]d@

    = (integral from 0 to 2pi/3) [ (3/4) + sqrt(3)cos@ + (cos@)^2 ]d@ - (integral from 2pi/3 to pi) [ (3/4) + sqrt(3)cos@ + (cos@)^2 ]d@

    = (integral from 0 to 2pi/3) [ (3/4) + sqrt(3)cos@ + (1/2)(1+cos2@) ]d@ - (integral from 2pi/3 to pi) [ (3/4) + sqrt(3)cos@ + (1/2)(1+cos2@) ]d@

    = [(3/4)@ + sqrt(3)*sin@ + (1/2)@ + (1/4)(sin2@) ] (integral from 0 to 2pi/3) - [(3/4)@ + sqrt(3)*sin@ + (1/2)@ + (1/4)(sin2@) ] (integral from 2pi/3 to pi)

    = [ (3/4)(2pi/3) + sqrt(3)*sin(2pi/3) + (1/2)(2pi/3) + (1/4)(sin( 2*(2pi/3) ) ] - (0 + 0 + 0 + 0) - [(3/4)(pi) + sqrt(3)*sin(pi) + (1/2)(pi) + (1/4)(sin( 2*(pi) ) ] + [ (3/4)(2pi/3) + sqrt(3)*sin(2pi/3) + (1/2)(2pi/3) + (1/4)(sin( 2*(2pi/3) ) ]

    = ( pi/2 + 3/2 + pi/3 - sqrt(3)/8 ) - ( 3pi/4 + pi/2 ) + ( pi/2 + 3/2 + pi/3 - sqrt(3)/8 )

    = 5pi/12 + 3 - sqrt(3)/4
     
  2. jcsd
  3. May 14, 2010 #2

    vela

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    Check your limits of integration.
     
  4. May 15, 2010 #3
    Thanks. The limits of integration was supposed to be 5pi/6 instead of 2pi/3.
     
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