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Triple Density Integration

  1. Nov 25, 2011 #1
    I’ve got two questions about two problems. First I just want to confirm that I’m setting up this density integral properly:

    “Find the mass of the conical solid bounded by [itex]z = \sqrt{x^2 + y^2}[/itex] and [itex]x^2 + y^2 + z^2 = 4[/itex] if the density at any point is proportional to the distance to the origin.

    I’m taking that last part to mean that Density = ρ, so when I convert to spherical coordinates and set up the integral I get:

    [itex] \int{0_2\pi} \int{0_\pi/4} \int{0_2} ρ^3 dρ d∅ dθ[/itex]

    Is that [itex] ρ^3 [/itex] right? Or should it be [itex] (ρ^2 + ρ) [/itex]?
     
  2. jcsd
  3. Nov 25, 2011 #2
    Okay, next question.

    “For the solid bounded by [itex] z = 4 - y^2[/itex], [itex]y = x[/itex], [itex]x = 0[/itex], [itex]z = 0 [/itex] find the volume using a double integral.”

    I did this using a triple, but I’m stuck trying to use a double. Converting to cylindrical coordinates, z goes from [itex]0[/itex] to [itex] 4 - r^2sin^2θ [/itex] while θ goes from ∏/4 to ∏/2. All well and good.

    But when I set up the integral:

    [itex] \int{\pi/2_\pi/4} \int{4 - r^2sin^2θ_0} r dr dθ [/itex]

    It’s clear that I’m going to end up with several unintegrated r terms. Gah, what am I missing?
     
  4. Nov 26, 2011 #3

    HallsofIvy

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    Use instead
    [ itex ] \int_0^{2\pi} \int_0^{\pi/4} \int_0^2 \rho^3 d\rho d\phi d\theta [/ itex ]
    to get
    [itex] \int_0^{2\pi} \int_0^{\pi/4} \int_0^2 \rho^3 d\rho d\phi d\theta [/itex]
    as far as the LaTeX goes.

    Have you thought about what you are doing? The mass of any object with density function [itex]\delta(x,y,z)[/itex] is [itex]\int\int\int \delta(x,y,z)dV[/itex]. I cannot imagine why you would think of adding.

    However, you have the differential of volume wrong. For spherical coordinates it is
    [tex]\rho^2 sin(\phi) d\rho d\phi d\theta[/tex]

    Also, saying that the density is proportional to the radius does not mean it is equal to the radius. It means it is equal to some constant times the radius.
     
  5. Nov 26, 2011 #4

    HallsofIvy

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    I cannot see any good reason to convert to cylindrical coordinates. There is no circular symmetry that woud make cylindrical coordinates simpler. In Cartesian coordinates, the integral for volume would be
    [tex]\int_{y= -2}^2\int_{z= 0}^{4- y^2}\int_{x= 0}^y dxdzdy[/tex]

    Can you convert that to a double integral?
     
  6. Nov 26, 2011 #5
    Thanks; this forum's code is very frustrating.

    Of course I've thought about it; that's why I'm here. If I hadn't thought about it, I'd have integrated according to my first instinct. Which in this case happened to be mostly right, but obviously I didn't know that.

    I've had no practice with this so let's find out...

    [itex] \int_0^{2} \int_0^{y} (4 - y^2) dx dy [/itex]
     
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