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Average Value and Area in Polar

  1. Dec 1, 2011 #1
    1. The problem statement, all variables and given/known data

    Let Q be the region bounded by r=sin(3θ) in the first quadrant. Find the area of Q. Find the average distance of points in Q from the origin.

    3. The attempt at a solution
    I thought I could calculate area like so:
    [tex]\int_0 ^{pi/3} \int_0 ^{sin(3θ)} sin(3θ) dr dθ[/tex] This gives answer [tex]π/6[/tex]But if I calculate area via a single integral, I have,[tex]\int_0 ^{pi/3} (1/2) (sin(3θ))^2 dθ = π/12[/tex]So clearly I am not doing something correctly in my calculation of the area.

    I'm not quite sure how to begin finding the average distance, but I assume I'll have to divide by the area at the end.
     
  2. jcsd
  3. Dec 1, 2011 #2

    Mark44

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    Your mistake is that you have sin(3θ) as both the integrand and as one of the limits of integration. Your integrand should be 1, and dA should be r dr dθ. With that change, after doing the inner integration, you get the same integral as below.
     
  4. Dec 1, 2011 #3

    Simon Bridge

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    What he said - also, I thought the first quadrant was 0-pi/2
     
  5. Dec 1, 2011 #4

    Mark44

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    It is, but as θ ranges from 0 to [itex]\pi[/itex]/3, 3θ ranges from 0 to [itex]\pi[/itex], hence r goes from 0 to 1 and then back to 0. Neither of the other two petals on this three-petaled rose is in the first quadrant, so the integration limits of 0 and [itex]\pi[/itex]/3 are fine.
     
  6. Dec 1, 2011 #5

    vela

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    r is the distance from the origin, so you want to calculate the average value of r. That's given by
    [tex]\langle r \rangle = \frac{1}{A} \int r\,dA[/tex]where A is the total area and dA is the area element Mark gave above.
     
  7. Dec 1, 2011 #6

    Simon Bridge

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    Ah - serves me right for not following my own advice and sketching it :)
     
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