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Finding the area enclosed by curves in polar form

  1. Jan 2, 2010 #1
    1. The problem statement, all variables and given/known data

    a) Find the area enclosed by the curve [tex]r=2+3cos(\theta)[/tex].

    b) Find the area enclosed by the curve [tex](x^2+y^2)^3=y^4[/tex] (after converting to polar form)

    2. Relevant equations

    The general equation for the area of a sector of curve:

    [tex] A=\frac{1}{2} \int_{\beta}^{\alpha} r^2 d\theta[/tex]

    3. The attempt at a solution

    I able to integrate part a) but unable to find the upper and lower limits to find the definite integral.

    For part b), likewise I am unable to find the limits in order to find the definite integral


    Thanks in advance,
    Charismaztex
     
  2. jcsd
  3. Jan 2, 2010 #2

    tiny-tim

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    Hi Charismaztex! Welcome to PF! :smile:

    (have a theta: θ and a pi: π and try using the X2 tag just above the Reply box :wink:)

    In polar coordinates, r can never be negative …

    so (in both cases) find the values of θ for which r = 0, and use them as your limits. :wink:

    (and if there's no "forbidden" values of θ, then just use limits of 0 to 2π)
     
  4. Jan 8, 2010 #3
    Hello, thanks for your input. I've solved part b) successfully but I am still stymied by the limits of part a). Solving for r=0 gives arccos(-2/3).

    -Charismaztex
     
  5. Jan 8, 2010 #4

    Mark44

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    Not true at all. The polar points (1, pi/4) and (-1, 5pi/4) refer to the same point, and clearly the r value in the 2nd coordinate pair is negative.
     
  6. Jan 8, 2010 #5

    LCKurtz

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    First you need to decide what are you are actually calculating. The graph is a limacon with an inner loop. If you just want the area enclosed by the limacon, you don't want to include the inner loop, which is what going from 0 to [itex]2\pi[/itex] would do. Next, you can notice from the symmetry you can just do the top half and double it.

    You have already noticed that angle where the top half is completed, minus the loop, is

    [tex]\alpha = \arccos{(-\frac 2 3) }[/tex]

    So if you let [itex]\theta[/itex] go from 0 to [itex]\alpha[/itex], you will get half the area. Now the fact that you can't get a nice expression for [itex]\alpha[/itex] shouldn't stop you. Just do the integral. Draw a picture of the angle [itex]\alpha[/itex] with a little triangle and you can read off the value of any trig function of [itex]\alpha[/itex] that comes out of your integration. You don't need to know the value of [itex]\alpha[/itex] itself.
     
  7. Jan 8, 2010 #6
    I'm confused about this question. Is anyone able to put up a full solution that would be appreciated. The textbook answer is [tex]5\sqrt{5} + \frac{17}{2} arccos(-\frac{2}{3}) [/tex]

    Thanks,
    Charismaztex
     
  8. Jan 8, 2010 #7

    vela

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    I think your book's answer is wrong. The first term should be [tex]3\sqrt{5}[/tex].
     
  9. Jan 8, 2010 #8

    LCKurtz

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    I agree.
     
  10. Jan 9, 2010 #9
    So that's the problem! The book apparently has many errors; I don't know if you've heard of it before, it's "Further Pure Mathematics Gaulter and Gaulter" published by Oxford. Thanks for the help, I got the answer.
     
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