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Homework Help: Triple integral in spherical coordinates

  1. Feb 29, 2012 #1
    I want to check if I'm doing this problem correctly.
    1. The problem statement, all variables and given/known data
    Region bounded by [itex]x^2+y^2=4[/itex] and bounded by the surfaces z = 0, and [itex]z=\sqrt{9-x^2-y^2}[/itex].
    Set up triple integrals which represent the volume of the solid using spherical coordinates.

    2. Relevant equations
    [itex]\int\int\int_{V}\rho^2sin\phi \; d\rho d\phi d\theta[/itex]

    3. The attempt at a solution
    The shape is a cylinder with a round top.
    It seems that I have to break this into two integrals:
    From [itex]x^2+y^2=4[/itex] and [itex]z=\sqrt{9-x^2-y^2}[/itex]:
    Since [itex]z=\rho cos\phi[/itex], [itex]\rho cos\phi = \sqrt{5} \Rightarrow \rho = \frac{\sqrt{5}}{cos\phi}[/itex]
    From [itex]z=\sqrt{9-x^2-y^2}[/itex], it is a sphere of radius 3 so ρ=3
    The sphere and cylinder meets at point (y,z)=(2, √5) and the radius makes an angle with the point:
    [itex]sin\phi = \frac{2}{3}[/itex]
    [itex]\int_{0}^{2\pi}\int_{0}^{arcsin(\frac{2}{3})}\int_{0}^{3}\rho^2sin\phi \; d\rho d\phi d\theta + \int_{0}^{2\pi}\int_{arcsin(\frac{2}{3})}^{\frac{ \pi }{2}}\int_{0}^{\frac{\sqrt{5}}{cos\phi}}\rho^2 sin \phi \; d\rho d\phi d\theta[/itex]
    Thank you.
  2. jcsd
  3. Feb 29, 2012 #2


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    Hello tintin2006. Welcome to PF !

    That all looks fine to me.
  4. Feb 29, 2012 #3
    Thanks :)
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