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Double Integral

  1. Apr 10, 2005 #1
    Greetings all,

    I need help setting up this problem:

    Use a double integral to find the area of the region enclosed by the curve

    r=4+3 cos (theta)

  2. jcsd
  3. Apr 10, 2005 #2

    1)You'll have to draw a polar curve to help you out with this question. From the drawn polar curve, you'll get the minimum value of r to be 1. (when cos(theta) is negative)
    2) Thus it follows that the range for r is 1<=r<=4+3cos(theta)
    Hence we'll integrate r from 1 to 4+3cos(theta)
    3) For the range of theta, you can use the range from 0 to pi for simplicity in calculations.(Just multiply the answer by 2 to obtain the full area.)

    I hope this helps =)
  4. Apr 10, 2005 #3


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    Yeah,a plot might help u convince why the limit wrt [itex] \theta [/itex] need to be 0 and [itex] \pi [/itex] and why you shouldn't integrate from 0 to [itex] 2\pi [/itex]

    I think it's a cardioide.

    Last edited: Apr 10, 2005
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