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Domain for Polar Coordinate Part 2

  1. Nov 13, 2011 #1
    1. The problem statement, all variables and given/known data
    f(x,y) = [itex]e^{x^2+y^2}[/itex]
    [itex]x^{2}[/itex] + [itex]y^{2}[/itex] ≤ R

    2. Relevant equations



    3. The attempt at a solution
    I believe this is a circle.

    [itex]f^{2pi}_{0}[/itex][itex]^{sqrt(R)}_{-sqrt(R)}[/itex][itex]e^{x^2+y^2}[/itex]*r*dr*dθ

    = [itex]f^{2pi}_{0}[/itex][itex]f^{sqrt(R)}_{-sqrt(R)}[/itex][itex]e^{r^2}[/itex]*r*dr*dθ

    after u substitution...

    = [itex]\frac{1}{2}[/itex][itex]f^{2pi}_{0}[/itex][itex]f^{R}_{R}[/itex][itex]e^{u}[/itex]*du

    Does this makes sense?

    Of course not... r domain are same...
     
  2. jcsd
  3. Nov 13, 2011 #2

    Dick

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    Polar coordinates are defined on the domain r>=0 and 0<=theta<2*pi. r is nonnegative. Your lower limit for r should be 0.
     
  4. Nov 13, 2011 #3
    ohhh so r can never be negative so they are always equal to or bigger than 0?
    ok I got it.
    Thank u

    So the correct one is

    = [itex]\frac{1}{2}[/itex][itex]f^{2pi}_{0}[/itex][itex]f^{R}_{0}[/itex][itex]e^{u}[/itex]*du
     
  5. Nov 14, 2011 #4

    HallsofIvy

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    Use \int for a [itex]\int[/itex] in LaTeX.
     
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