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Triple integral in spherical coordinates

  1. Nov 7, 2012 #1
    1. The problem statement, all variables and given/known data

    The problem is to calculate the volume of the region contained within a sphere and outside a cone in spherical coordinates.

    Sphere: x2+y2+z2=16

    Cone: z=4-√(x2+y2)

    2. Relevant equations

    I am having difficulty converting the equation of the cone into spherical coordinates. Judging by the graph I was able to deduce from the formulas (working in rectangular coordinates), I believe I will need the cone equation as the inside ρ limit.

    3. The attempt at a solution

    Converting the sphere into spherical coordinates:

    x2+y2+z2=16

    ρ2=16

    ρ=4 a sphere with radius 4 centered at the origin, which is consistent with my graph.

    I recognize that the cone is downward opening and peaks at z=0. My attempt to convert the equation was as follows:

    ρcos[itex]\phi[/itex]=4-√(ρ2sin2[itex]\phi[/itex]cos2θ+ρ2sin2[itex]\phi[/itex]sin2θ)

    ρcos[itex]\phi[/itex]=4-ρsin[itex]\phi[/itex]

    ρ(cos[itex]\phi[/itex]+sin[itex]\phi[/itex])=4

    I can't figure out how to further simplify this formula. All the examples of cones in spherical coordinates I came across were peaked at the origin and simplified nicely to [itex]\phi[/itex]=(some arbitrary angle), but I couldn't find any that were more complicated.

    This is my first post so please let me know if I've done anything wrong, and thanks in advance!
     
  2. jcsd
  3. Nov 8, 2012 #2

    tiny-tim

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    welcome to pf!

    hi dickyroberts! welcome to pf! :smile:
    i'm not surprised! :rolleyes:

    just because it's a sphere, that's no reason to jump straight in and convert to spherical coordinates! :wink:

    it has no spherical symmetry, so spherical coordinates just makes it more complicated :redface:

    you could use cylindrical coordinates (since it obviously has cylindrical symmetry),

    or it might be easier to stick to Cartesian coordinates (since you already have them), and slice the region into horizontal slices of height dz …

    try both ways :smile:

    btw …
    … (you've probably worked this out by now, but anyway:) no, it peaks at z = 4 :wink:
     
  4. Nov 8, 2012 #3

    HallsofIvy

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    tiny-tim, my interpretation of the problem is that we are required to use spherical coordinates.

    dickeyroberts, you say you got [itex]\rho(cos(\phi)+ sin(\phi))= 4[/itex]. Okay, what do you get with [itex]\rho= 4/(cos(\phi)+ sin(\phi))[/itex]?
     
  5. Nov 8, 2012 #4
    Hey tiny-tim, thanks for the welcome!

    and my mistake, that was a typo. I meant to say the cone peaks at z=4, that wouldn't have been a very good start :wink:

    HallsofIvy - I think you may have just solved my problem!

    For some reason I had the idea that I needed the equation of the cone to be ϕ=(arbitrary angle) because that's what all the examples I could find had. Not sure what I was thinking; your last comment helped me realize that I can just use the equation

    ρ=4/(cos(ϕ)+sin(ϕ)

    as the limit in the integration of ρ since ϕ will be integrated in a later step.

    The biggest downside of doing an online course is not having classmates to bounce ideas off of :rolleyes:

    Thanks guys!
     
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