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Convert to cylindrical coordinates

  1. Nov 15, 2009 #1
    Evaluate by changing to cylindrical coordinates

    [tex]\int[/tex] from 0 to 1 [tex]\int[/tex] from 0 to (1-y^2)^1/2 [tex]\int[/tex] from (x^2+y^2) to (x^2+y^2)^1/2 (xyz) dzdxdy

    I came to an answer of integral from 0 to pi integral from 0 to 1 integral from r^2 to r (rcos[tex]\theta[/tex]rsin[tex]\theta[/tex]z) r dzdrd[tex]\theta[/tex]
    Is this the correct answer?
     
  2. jcsd
  3. Nov 15, 2009 #2

    LCKurtz

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    Hello Caliguy. Click on the expression below to see how to post it so it is readable in tex:

    [tex]\int_0^1 \int_0^{\sqrt{1-y^2}}\int_{x^2+y^2}^{\sqrt{x^2+y^2}}xyz\ dzdxdy[/tex]

    This looks like a first octant integral. Check your [itex]\theta[/itex] limits.
     
  4. Nov 15, 2009 #3
    To me they seem right, doesn't theta go from zero to pi? after graphing it it looks like a half a circle... maybe I'm overlooking something?
     
  5. Nov 15, 2009 #4

    LCKurtz

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    Neither x nor y get negative in your original integrals.
     
  6. Nov 15, 2009 #5
    So theta only goes from 0 to pi/2 right?
     
  7. Nov 15, 2009 #6

    LCKurtz

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    Yes.
     
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