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Homework Help: Integration of Polar coordinates

  1. Jan 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the area in the polar curve r = sin2θ between 0 and [itex]\frac{\pi}{2}[/itex].

    The way to do this is to say the area of a tiny bit of this polar curve, dA = [itex]\frac{1}{2}[/itex]r[itex]^{2}[/itex]dθ

    so the integral is just [itex]\frac{1}{2}[/itex][itex]\int[/itex][itex]^{\frac{\pi}{2}}_{0}[/itex](sin2θ)[itex]^{2}[/itex]dθ

    if we did say a function in cartesian coordinates, eg y=x we just do [itex]\int[/itex]x dx. I am confused as to why in polar coordinates we cannot just do the same and say do [itex]\int[/itex]sin2θ dθ, I understand how to get the correct integral, but what I am asking is what is wrong with just integrating the polar function like we do in cartesian coordinates? why do we have to say we will take a bit of this sector and integrate it over these limits, for polar function, but for a cartesian one we just plug the function into the integral?

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

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jan 8, 2013 #2


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    Staff: Mentor

    In both methods, you skip one step to calculate an area. In cartesian coordinates, this step is trivial (you have to multiply with 1), in polar coordinates, it is not.
    See the discussion here, for example.
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