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Changing a double integral to polar coordinates

  1. Oct 15, 2009 #1

    Ral

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    1. The problem statement, all variables and given/known data
    Rewrite by converting to polar coordinates, carefully drawing R.

    [tex]\int^{2}_{0}\int^{\sqrt{2x-x^2}}_{0}\sqrt{x^2+y^2}dydx[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I believe I have the inside part of it right. What I did was replace the x^2 and y^2 in [tex]\sqrt{x^2+y^2}[/tex] with [tex]r^2cos^2\vartheta[/tex] and [tex]r^2sin^2\vartheta[/tex], factored out the r^2, which left me with [tex]cos^2\vartheta+sin^2\vartheta[/tex], which equals 1. So then I would be left with [tex]\sqrt{r^2}rdrd\vartheta[/tex]

    What's confusing me is what my limits on the integrals should be. What I have right now is

    [tex]\int^{\pi/2}_{0}\int^{2}}_{0}\sqrt{r^2}rdrd\vartheta[/tex]
    but I'm still unsure if that's correct or not.
     
    Last edited: Oct 15, 2009
  2. jcsd
  3. Oct 15, 2009 #2

    Mark44

    Staff: Mentor

    Have you drawn the region R? From your Cartesian integral, the limits for y range from 0 to sqrt(x^2 - 4). You show this as sqrt(x^2 - 2^2). Is there some reason you wrote 2^2 instead of 4?

    Can you show us your reasoning in changing the limits of integration as you did?

    One last thing. In your polar integral you should simplify sqrt(r^2) to r.
     
  4. Oct 15, 2009 #3

    Ral

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    That limit should have been [tex]\sqrt{2x-x^2[/tex]. I messed up typing it up, I edited it in the right limit.
     
  5. Oct 15, 2009 #4

    Mark44

    Staff: Mentor

    OK, I thought there might be a typo there. Your region R over which integration takes place is bounded by y = 0 and y = sqrt(2x - x^2), and by x = 0 and x = 2.

    Can you draw this region? This problem also asks you to do that. The regions is a simple geometric shape, and one that will make the polar integral very easy.

    Many times in these types of problems, the hardest part is understanding exactly what the region of integration looks like. If you can't do this, changing the limits will pretty much be insurmountable.
     
  6. Oct 15, 2009 #5

    Ral

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    http://img96.imageshack.us/img96/5444/66431931.jpg [Broken]

    Here's what I had. The 1st quadrant would be shaded. The one that's confusing me the most is the integral for dr, while I think the integral for d[tex]\vartheta[/tex] is correct.
     
    Last edited by a moderator: May 4, 2017
  7. Oct 15, 2009 #6

    Mark44

    Staff: Mentor

    Yes, that's what R looks like.
    Do you mean the limits of integration for r and [tex]\vartheta[/tex]?
    Otherwise, I don't understand what you mean.
     
  8. Oct 15, 2009 #7

    Ral

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    Yes, I mean the limits of integration for r and theta. So if that R is correct, then that means I had the correct answer, right?
     
  9. Oct 15, 2009 #8

    Mark44

    Staff: Mentor

    Yes, but I would write the polar integral as [tex]\int^{\pi/2}_{0}\int^{2}}_{0}r^2drd\vartheta[/tex]

    In the polar interpretation of R, r extends from 0 to 2. All points on the circle are 2 units from its center. [itex]\theta[/itex] extends from 0 to [itex]\pi[/itex]/2. That gives you the quarter circle in the first quadrant.
     
  10. Oct 15, 2009 #9

    Ral

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    Yeah, I was gonna do that simplification. Thanks for helping me clarify this.
     
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