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Check this question to see if it is right.

  1. Oct 5, 2007 #1
    1. The problem statement, all variables and given/known data
    Find area between the polar graph[tex] r= \theta[/tex] and [tex] r=2\theta[/tex]
    from [tex] 0 \leq \theta \leq 2\pi [/tex].

    2. Relevant equations
    [tex] Area = \frac{1}{2} \int r^2 \theta [/tex]

    3. The solution

    [tex] Area = \frac{1}{2} \int_0^2 (2 \theta)^2 d\theta - \frac{1}{2} \int_0^2 \theta^2 d\theta [/tex]

    final answer [tex] 4\pi^3 [/tex] or 124.025

    check please as soon as possible, only have to say right or wrong!
    Last edited: Oct 5, 2007
  2. jcsd
  3. Oct 5, 2007 #2
    I put the data to mathematica by using the integrals you wrote and i got 1.0472. I dont understand how you derived the integral and what exactly you ask. Please re write the question so we can understand better what you want.
  4. Oct 5, 2007 #3
    looks wrong,

    for a polar area your integral should be:
    [tex]\frac{1}{2}\int_{a}^{b}r^2 d\theta[/tex]

    also your a/b seem wrong, it should be 0 and 2pi but you have 0 and 2
  5. Oct 5, 2007 #4
    sorry i type some error, I edit the question above!
    check again!
  6. Oct 5, 2007 #5
    is from 0 to 2pi for the integrals
  7. Oct 5, 2007 #6
    According to your integrals and to your latest corrections the value you get is correct. I double checked with mathematica and the answer is ok.
  8. Oct 5, 2007 #7
    yep that answer seems correct now
  9. Oct 5, 2007 #8
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