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Area of Polar Graph

  1. May 10, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the area of the region.
    Interior of: r = 2 - sin(b)

    2. Relevant equations

    A = 1/2 ∫ (r)^2 dr

    3. The attempt at a solution

    I really don't have any idea how to approach this problem. I don't understand how to determine my limits of integration. The only part of the problem I have accomplished is finding that b = arcsin(2). Also, I have found that the graph of this equation makes a convex limacon.
  2. jcsd
  3. May 10, 2014 #2


    Staff: Mentor

    You should be able to figure out the integration limits from the graph. You could integrate from b = 0 to b = ##2\pi##. Alternatively, you could integrate from b = ##\pi/2## to b = b = ##3\pi/2## and double that result.
  4. May 10, 2014 #3
    ok so would my equation look something like ∫ [(1-sin(b))^2 db] with the limits of integration going from 0 to 2pi?
  5. May 10, 2014 #4


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    Looks good, provide [itex]b[/itex] is the polar angle. Why don't you use LaTeX for typesetting formulas? It's no big deal to type it and it makes everything much easier to read and understand!
  6. May 10, 2014 #5
    I don't know how to use latex, nor do I know where to find it....sorry
  7. May 10, 2014 #6
    Ok, I am confused about this problem now. The website for my calculus textbook offers a 24/7 tutoring service. When I asked the tutor about this problem he said I could not integrate from 0 to 2pi; however, he was also not sure how to do the problem. So can I safely assume that he was wrong about being unable to integrate from 0 to 2pi?
  8. May 10, 2014 #7


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    Well, could you post the full question? I don't know, why one shouldn't integrate over the full range of the angle, because [itex]r(b)=2-\sin b>0[/itex] for all [itex]b \in [0,2 \pi][/itex].
  9. May 10, 2014 #8


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    Yes, like this:
    Code (Text):

    ##\frac 1 2\int_{0}^{2\pi} (2-\sin\theta)^2d\theta##
    which displays like this: ##\frac 1 2\int_{0}^{2\pi} (2-\sin\theta)^2d\theta##
  10. May 10, 2014 #9


    Staff: Mentor

    It doesn't give me a lot of confidence when someone tells me I can't do a problem a certain way, but then doesn't know how to do the problem.
    If you look at the graph, it's pretty obvious that you can integrate using those limits, regardless of what the tutor is saying.
  11. May 10, 2014 #10
    I did post the full question.
  12. May 10, 2014 #11
    Thank you...my first instinct was to integrate from 0 to 2pi; however, the tutor swayed me lol...
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