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Area in Polar Coordinates

  1. Jul 29, 2017 #1
    Hi, everyone. I had an example from my book, but I wasn't sure how they got [itex]\dfrac{1}{2}cos\theta[/itex] on step 7? It seems like once they combined the constants, they ended up with just [itex]cos2\theta[/itex]. Although, they have a [itex]\dfrac{1}{2}[/itex] in front. Can someone help me understand where that constant came from? Thank you.

    1. The problem statement, all variables and given/known data

    Find the area of the region enclosed by the polar curve [itex]r=3-3cos\theta[/itex]

    2. Relevant equations

    Area in Polar Coordinates
    [itex]dA=\dfrac{1}{2}r^2d\theta[/itex]

    Double Angle Identity:
    [itex]cos2\theta=2cos^2\theta-1[/itex]

    3. The attempt at a solution

    1. [itex]dA=\dfrac{1}{2}(3-3cos\theta)^2d\theta[/itex]

    factor

    2. [itex]dA=\dfrac{1}{2}(9-18cos\theta+9cos^2\theta)d\theta[/itex]

    write the integral and factor [itex]\dfrac{1}{2}[/itex] through

    3. [itex]\displaystyle A=\dfrac{1}{2}\int_{0}^{2\pi}(9-18cos^2\theta+9cos^2\theta)d\theta[/itex]

    factor out 9

    4. [itex]\displaystyle A=9\cdot\dfrac{1}{2}\int_{0}^{2\pi}(1-2cos\theta+cos^2\theta)d\theta[/itex]

    rewrite double-angle identity

    5. [itex]cos^2\theta=\dfrac{cos^2\theta+1}{2}[/itex]

    replace [itex]cos^2\theta[/itex] with the double-angle identity

    6. [itex]\displaystyle A=\dfrac{9}{2}\int_{0}^{2\pi}\left(1-2cos\theta+\dfrac{cos^2\theta+1}{2} \right)d\theta[/itex]

    combine the constants to get [itex]\dfrac{3}{2}[/itex]

    7. [itex]\displaystyle A=\dfrac{9}{2}\int_{0}^{2\pi}\left(\dfrac{3}{2}-2cos\theta+\dfrac{1}{2}cos2\theta \right)d\theta[/itex]

    integrate

    8. [itex]\dfrac{9}{2}\left[\dfrac{3}{2}\theta-2sin\theta+\dfrac{1}{4}sin\theta \right]_{0}^{2\pi}[/itex]

    answer

    9. [itex]\dfrac{27\pi}{2}[/itex]
     

    Attached Files:

  2. jcsd
  3. Jul 29, 2017 #2

    blue_leaf77

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    Check again the double angle identity you used in step 5.
     
  4. Jul 29, 2017 #3
    I meant to write [itex]\dfrac{cos2\theta+1}{2}[/itex] for step 5 and 6. Although I do have it written for step 7.
     
  5. Jul 29, 2017 #4

    blue_leaf77

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    So are the steps above the one given in your book or your own? I don't see any flaw in it except for the typo in writing the double angle identity.
     
  6. Jul 29, 2017 #5
    Not my own. From the book. I just don't get where the[itex]\dfrac{1}{2}[/itex] comes from in step7?
     
  7. Jul 29, 2017 #6

    blue_leaf77

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    $$
    \cos^2\theta=\dfrac{\cos 2\theta+1}{2} = \frac{1}{2}\cos 2\theta + \frac{1}{2}$$
     
  8. Jul 29, 2017 #6

    Ray Vickson

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    Just for the record: don't write ##cos 2 \theta## in LaTeX; it looks ugly and is hard to read. Write ##\cos 2 \theta## instead. You do that by typing "\cos" instead of "cos", and that goes for most other elementary functions as well.
     
  9. Jul 29, 2017 #7
    Alright, noted.

    Oh, you split up the [itex]\dfrac{\cos2\theta+1}{2}[/itex] into two denominators like [itex]\dfrac{\cos2\theta}{2}+\dfrac{1}{2}[/itex] That makes sense, thanks.
     
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