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Converting to polar double integrals

  1. May 12, 2009 #1
    1. The problem statement, all variables and given/known data
    we are given the intgral from 0 to 2( the integral from 0 to sqrt(1-(x-1)^2) of ((x+y)/(x^2+y^2))dydx, so convert to polar integral and solve


    2. Relevant equations



    3. The attempt at a solution

    i got integral from 0 to pi/2(integral 0 to 2cos(theta) of (x+y)/r dr d(theta), and not sure if im right and dunno how to procede. i let x=1+cos(theta) and y=sin(theta) and r^2=x^2+y^2. sorry bout the ugly format but i dunno how to write it out in maths form.

    Thanks in advanced
     
    Last edited: May 12, 2009
  2. jcsd
  3. May 12, 2009 #2

    HallsofIvy

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    Strictly speaking, you have shifted the coordinate system so the origin is at the old (1, 0). That perfectly valid but might violate instruction to just change to polar coordinates.

    If [itex]x= 1+ cos(\theta)[/itex] and [itex]y= sin(\theta)[/itex] then [itex]\sqrt{1- (x-1)^2}= \sqrt{1- r^2cos^2(\theta)}= sin(\theta)[/itex] so your equation becomes [itex]y= r sin(\theta)= sin(\theta)[/itex] and [itex]r= 1[/itex], a circle about the origin with radius 1 (the graph in the original coordinates is, of course, a circle about (1, 0) with radius 1).

    But the fact that y is equal to the positive square root means you are only getting the upper half of the circle. The integral is from [itex]\theta= 0[/itex] to [itex]\pi[/itex], not [itex]2\pi[/itex]. And, in polar coordinates, the "differential of area", dxdy, becomes [itex]rdrd\theta[/itex].

    Your integrand is [itex](x+y)/(x^2+ y^2)= [/itex]
    [tex]\int_{r=0}^1\int_{\theta= 0}^\pi (r cos(\theta)+ r sin(\theta))/r^2= (1/r)(cos(\theta)+ sin(\theta)) r dr\theta[/itex]
     
    Last edited: May 13, 2009
  4. May 12, 2009 #3
    ok i see, i thought about moving the orgin but i thought that would change my values of (x+y)/(x^2+ y^2) when integrating and screw up my integral? So in general, can i move the origin and the function in the double integral (in this case (x+y)/(x^2+ y^2) ) will always fix itself?
     
  5. May 12, 2009 #4
    okay i did your double integral of sin(theta) + cos(theta) drd(theta) and i got 2 yet the answers in the book have pi/2+1.

    i tried letting x=1+rcos(theta) and y=rsin(theta), r^2=x^2+y^2, and i get the double integral of 1/r+cos(theta)+sin(theta), but when i antiderive with respect to r i get log|r|+... and when i sub in the terminals for r(0 and 1) i get log|0| which is undefined.

    any help please?
     
  6. May 14, 2009 #5
    Good question Cos(e).

    Worked Solution attached as Word Document. Fingers Crossed the file opens.
     

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