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Polar Coordainates Question

  1. Oct 18, 2009 #1
    Find the area enclosed by the curve r=(1+Cos[tex]\theta[/tex]),0[tex]\leq[/tex] [tex]\theta[/tex] [tex]\leq[/tex] 2[tex]\pi[/tex]



    |cos [tex]\theta[/tex]|[tex]\leq[/tex]1
    Maximum r =2(1+1)=4

    When r=0,
    2+2Cos[tex]\theta[/tex]=0
    Cos[tex]\theta[/tex]=-1
    Key angle=0
    [tex]\theta[/tex]=[tex]\pi[/tex],[tex]3/2\pi[/tex]

    Area of curve =1/2[tex]\int[/tex]r^2 d[tex]theta[/tex]
     
    Last edited: Oct 18, 2009
  2. jcsd
  3. Oct 18, 2009 #2

    CompuChip

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    So what did you try already?
     
  4. Oct 18, 2009 #3
    I dont know how to sketch the graph?!?
     
  5. Oct 18, 2009 #4

    lanedance

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    try setting up the area integral and we can discuss

    here's a hint & some example tex for you try clicking below (another hint)
    hint: as a start what is an infintesimal area element in polar coordinates?
    [tex] \int_0^{2 \pi} d \theta \int_0^{r(\theta)} dr f(r,\theta) [/tex]
     
  6. Oct 18, 2009 #5

    HallsofIvy

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    When [itex]\theta= 0[/itex], r= 1+1= 2 so one point on the graph is at (2,0) When [itex]\theta= \pi/2[/itex], r= 1+ 0= 1 so another point is at (0, 1). When [itex]\theta= \pi[/itex], r= (1+ (-1))= 0 so a third point is (0,0). When [itex]\theta= 3\pi/2[/itex], r= 1+ 0= 1 so a fourth point is (0, -1). When [itex]\theta= 2\pi[/itex] r= 1+1= 2[/itex] so we are back at (2, 0). That's all you need to know: we go around one complete circuit of the figure as [itex]\theta[/itex] goes from 0 to [itex]2\pi[/itex]. You don't need to actually draw the graph. (It is a figure known as a "cardioid".)
    Integrate [itex]rd\theta= (1+ cos(\theta)d\theta[/itex] from [itex]\theta= 0[/itex] to [itex]\theta= 2\pi[/itex].
     
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