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Homework Help: Converting into polar coordinates for integration

  1. Apr 11, 2005 #1
    Another problem that I cannot figure out. Convert the follorwing into polar coordinates:

    [tex]\int_{-1}^{1} \int_{-\sqrt{1 - y^2}}^{\sqrt{1 - y^2}} ln\left(x^2 + y^2 + 1\right) dx\;dy[/tex]

    I did this so far:

    [tex]ln\left(x^2 + y^2 + 1\right) = ln\left(r^2 + 1\right)[/tex]

    [tex]\sqrt{1 - y^2} = \sqrt{1 - r^2 \sin^2 \theta}[/tex]

    Now what do I do?

    Is this possibly right?

    [tex]\int_{0}^{2\pi} \int_{-\sqrt{1 - r^2 \sin^2 \theta}}^{\sqrt{1 - r^2 \sin^2 \theta}} ln\left(r^2 + 1\right) dr\;d\theta[/tex]
     
    Last edited: Apr 11, 2005
  2. jcsd
  3. Apr 11, 2005 #2
    Trig identities, what does [itex] 1-sin^2(\theta) [/tex] equal?
     
  4. Apr 11, 2005 #3
    Better yet,

    x^2+y^2 = r^2
    with r = 1,
    x^2+y^2 = 1
    x^2 = 1-y^2

    x = sqrt(1-y^2)
     
  5. Apr 11, 2005 #4

    SpaceTiger

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    If what whozum said isn't clear to you, try drawing a graph of the region you're integrating over. Also, remember that the area element for a polar integral is

    [tex]dA=rdrd\theta[/tex]
     
  6. Apr 11, 2005 #5
    Is this it?

    Including the [tex]dA = r\;dr\;d\theta[/tex], is this integral the right conversion?

    [tex]\int_{0}^{2\pi} \int_{-\sqrt{1 - r^2 \sin^2 \theta}}^{\sqrt{1 - r^2 \sin^2 \theta}} r\;ln\left(r^2 + 1\right) dr\;d\theta[/tex]
     
  7. Apr 11, 2005 #6
    The limits of r simplify, other than that your right. Look at our hints above.
     
  8. Apr 11, 2005 #7
    Cool , but how does the [tex]1-sin^2(\theta) = \cos^2\left(\theta\right)[/tex] help when I have these limits of integration for [itex]r[/itex]?

    [tex]\sqrt{1 - r^2 \sin^2 \theta}}[/tex]

    It does not factor right?

    Attached is a graph.
     

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    Last edited: Apr 11, 2005
  9. Apr 11, 2005 #8

    SpaceTiger

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    Does it make sense to have "r" in the limit of an integral over r?
     
  10. Apr 11, 2005 #9
    It does factor, thats my point. Read what I said above clearly, sketch the domain. It will be clear what your limits for R are.
     
  11. Apr 11, 2005 #10
    Let's try this?

    [tex]x = \sqrt{1 - y^2}[/tex]
    [tex]r\;\cos\theta = \sqrt{1 - r^2\;\sin^2\theta}[/tex]

    Then solve for r?
     
  12. Apr 11, 2005 #11
    Like this?

    [tex]r^2\;\cos^2\theta = 1 - r^2\;\sin^2\theta[/tex]

    [tex]r^2\;\left(\cos^2\theta + \sin^2\theta\right) = 1[/tex]

    [tex]r^2 = 1[/tex]

    [itex]r[/itex] limits are then from [itex]0[/itex] to [itex]1[/itex]?

    The right integral?

    [tex]\int_{0}^{2\pi} \int_{0}^{1} r\;ln\left(r^2 + 1\right) dr\;d\theta[/tex]
     
  13. Apr 11, 2005 #12
    Correct, though it might be -1 to 1, spacetiger will give you the final ok.
     
  14. Apr 11, 2005 #13

    HallsofIvy

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    The point of all the comments was : DRAW A PICTURE!
    [tex]y= \sqrt{1- x^2}[/tex] and [tex]y=-\sqrt{1- x^2}[/tex] both are the same as [tex]x^2= 1- y^2[/tex] or [tex]x^2+ y^2= 1[/tex], the unit circle. Since x ranges from -1 to 1, the region of integration is simply the unit circle. To cover that, [tex]\theta[/tex] ranges from 0 to [tex]2\pi[/tex] and r from 0 to 1.
     
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