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Homework Help: Triple integral xyz coordinates to spherical

  1. Apr 27, 2010 #1
    1. The problem statement, all variables and given/known data

    the integral is bounded below by the cone [tex]z^2=x^x+y^2[/tex] and above by the sphere [tex]x^2+y^2+z^2=18[/tex]

    [tex]\int^{3}_{0}\int^{\sqrt{9-y^2}}_{0}\int^{\sqrt{18-x^2-y^2}}_{\sqrt{x^2+y^2}}x^2+y^2+y^2 dzdxdy[/tex]

    2. Relevant equations

    conversion formulas

    3. The attempt at a solution

    I am struggling on my limits. Obviously these limits are not true
    [tex]\int^{3}_{0}\int^{\sqrt{9-y^2}}_{0}\int^{\sqrt{18-x^2-y^2}}_{\sqrt{x^2+y^2}}p^2 p^2sin\phi dp d\theta d\phi [/tex]


    I know the upper limit for p is going to be [tex]\sqrt{18}[/tex] as that is the upper bound of the sphere but the lower bound of p is the cone [tex]z^2=x^2+y^2[/tex]

    is this [tex]x^2+y^2-z^2 = (psin\phi cos\theta )^2 + (psin\phi sin\theta )^2 - (pcos\phi)^2 [/tex] for the lower limit of p when factoring out to get p = something??? It seems a little long...

    //Edit

    ok, i had another look at this

    Is my domain as followed:

    D = {[tex](\theta , \phi \ , \rho) | (0 \leq \theta \leq 2\pi , 0 \leq \phi \leq \frac{\pi}{4} , 0 \leq \rho \leq \sqrt{18}[/tex]}

    Is this correct for a cone within a sphere? Im not too certain about [tex]\rho[/tex]
     
    Last edited: Apr 28, 2010
  2. jcsd
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