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Double integral in polar form

  1. Jul 9, 2016 #1
    1. The problem statement, all variables and given/known data
    can someone explain about the formula of the circled part?
    Why dA will become r(dr)(dθ)?
    MhTV4wv.png
    2. Relevant equations


    3. The attempt at a solution
    A = pi(r^2)
    dA will become 2(pi)(r)(dr) ?
    why did 2(pi) didnt appear in the equation ?
     
  2. jcsd
  3. Jul 9, 2016 #2

    SteamKing

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    dA = r ⋅ dr ⋅ dθ as can be seen from the diagram below:


    polar_coordinates_area_calculation.png

    2π has nothing to do with converting dA into its polar equivalent.
     
  4. Jul 10, 2016 #3

    vela

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    The dA you calculated here is the infinitesimal increase in the area of a circle if you increase the radius by dr. In other words, it's the area of a ring of radius r and width dr. It's not the same dA that appears in the integral.
     
  5. Jul 26, 2016 #4

    BvU

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    Is it clear to you now that this hasn't happened because the integral reads$$
    \displaystyle \iint\limits_R f(x,y)\; dA = \iint\limits_R f(r,\theta)\; r \;dr\;d\theta\quad ?$$Only if ##\ f(r,\theta) = f(r)\ ## i.e. f does not depend on ##\theta##, the integration over ##d\theta## can be carried out (yielding ##2\pi##) and the result is$$
    \displaystyle \iint\limits_R f(x,y)\; dA = \int\limits_{r_0}^{r_1} f(r)\; r \;dr\;\int\limits_0^{2\pi} d\theta\quad = 2\pi \int\limits_{r_0}^{r_1} f(r)\; r \;dr\;$$
     
  6. Jul 26, 2016 #5
    yes, thanks
     
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