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Homework Help: Finding the area enclosed by curves in polar form

  1. Jan 10, 2010 #1
    1. The problem statement, all variables and given/known data

    a)
    [tex]r=a(2+ cos(\theta))[/tex]
    Find the area of the region enclosed by the curve giving answers in terms of [tex]\pi[/tex]
    and [tex]a[/tex]

    b) Show that the area enclosed by the loop [tex]r=2(1-sin(\theta))\sqrt{cos(\theta)}[/tex] is [tex]\frac{16}{3}[/tex] and show that the initial line divides the area enclosed by the loop in the ratio 1:7 [tex](\theta[/tex] is between [tex]-\frac{\pi}{2} and \frac{\pi}{2})[/tex]
    2. Relevant equations

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

    3. The attempt at a solution

    a) I tried using [tex]\theta=0, \pi, \frac{\pi}{2}, \frac{3\pi}{2}[/tex] as limits and adding the quadrants up but comes out to be the incorrect answer. (textbook answer: [tex]\frac{9a^2\pi}{2}[/tex])

    b) I'm having a hard time integrating the [tex] cos(\theta)cos(2\theta)[/tex] part after squaring r. I know that one of the limits is [tex]\frac{\pi}{2}[/tex] but I'm not sure what the other should be.

    Thanks for your help,
    Charismaztex
     
  2. jcsd
  3. Jan 10, 2010 #2

    LCKurtz

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    For (a) your limits should be from 0 to [itex]2\pi[/itex].

    For (b), since [itex]\cos\theta[/itex] is negative on part of its domain, you want to use the interval [itex][-\pi/2,\pi/2][/itex]. I don't see any need for a [itex]\cos{2\theta}[/itex]. When you square r, you get some sines times a cosine, so a u substitution should work.
     
  4. Jan 10, 2010 #3
    So what would be the best way to integrate [tex]sin^2(\theta)cos(\theta)[/tex]?
     
  5. Jan 10, 2010 #4

    LCKurtz

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    u substitution. Try [itex]u = \sin\theta[/itex]
     
  6. Jan 10, 2010 #5
    Ahh yes, that worked fine! so I've got the area to be[tex]\frac{16}{3}[/tex], what intervals do you recommend (the part I normally struggle with) to calculate the separate area in order to get the ratio?
     
  7. Jan 10, 2010 #6

    LCKurtz

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    The problem itself says "show the initial line" divides the area. I take that to mean the x axis. Try it.
     
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