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Cartesian to polar

  1. Nov 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral.
    int(-1to1)int((sqrt(1-y^2))to(sqrt(1-y))[x^2+y^2]dxdy

    2. Relevant equations
    x=rcostheta
    y=rsintheta


    3. The attempt at a solution
    int(-1to1)int((sqrt1-(rsintheta)^2)to(sqrt(1-(rsintheta))))[(rcostheta)^2+(rsintheta)^2]rdrdtheta
     
  2. jcsd
  3. Nov 17, 2009 #2

    CompuChip

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    Note that [x^2+y^2] is simply r^2 (by definition!).

    Also you cannot just change your integration boundaries like that. For example, neither theta nor r runs between -1 and 1.
    I suggest drawing an image of the integration region, then think about how to describe it in terms of boundaries on r and theta.
     
  4. Nov 17, 2009 #3
    ok so i am not sure how to get the new boundries
    but its r^2rdrdtheta which integrated goes to r^4/4dtheta. Is this right?
     
  5. Nov 17, 2009 #4

    HallsofIvy

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    Yes, [itex]\int r^3dr[/itex] is [itex]r^4/4[/itex].

    Are you sure about that "[itex]y= \sqrt{1- x}[/itex] limit? That will give the right part of a parabola for y> 0 it is inside the unit circle given by the lower limit and for y< 0, it is outside.
     
  6. Nov 17, 2009 #5
    i thought it was x=sqrt(1-y)
     
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