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Ok another few problems

  1. Jul 12, 2005 #1


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    I really shouldnt be stuck on those problems but for whatever reason i cant solve them

    Problem: A particular spherical cloud of gas of radius 3 km is more dense at
    the center than towards the edge. The density, D, of the gas at a distance p km
    from the center is given by [tex]D(p) = 9 - p^2[/tex]. Write an integral representing the total mass of the cloud of gas, and evaluate it.

    Solution: density = mass/area. The spherical cloud's area is 9pi/2
    mass =9/2 pi (9-p^2).

    So mass = [tex]\frac{9pi}{2} \int_{0}^{pi} \int_{0}^{3} (9-r^2) \ r \ dr \ dtheta [/tex]

    Is this correct?
  2. jcsd
  3. Jul 12, 2005 #2
    If you are looking for a spherical cloud of gas you should have an integral in 3 dimensions instead of an integral in two dimensions. It should be a triple integral. Also where did the term 9pi / 2 come from?

    What you want to do is find a triple integral that will give you the total area of the sphere and count the density function over that volume.

    Last edited: Jul 12, 2005
  4. Jul 12, 2005 #3


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    Volume of spherical cloud with r=3 ->> V=(4/3)*pi^3= 4pi/3

    [tex]mass \ = \ \frac{4pi}{3} \ \int_{0}^{pi} \int_{0}^{2pi} \int_{0}^{3} (9-p^2) p^2 \ sin(phi) \ dp \ dtheta \ dphi[/tex]
  5. Jul 13, 2005 #4
    close, but you don't need the factor of 4*pi/3 there, the 4pi/3 factor will drop out of the integrals as such:

    for a circle of radius=1

    [tex]volume\ = \int_{0}^{\pi} \int_{0}^{2\pi} \int_{0}^{1}p^2 \ sin(\phi) \ dp \ d\theta \ d\phi[/tex]

    [tex]volume\ = 2\pi \int_{0}^{\pi} \int_{0}^{1}p^2 \ sin(\phi) \ dp d\phi[/tex]

    [tex]volume\ = 2\pi \frac{1}{3}p\right)^{3} \ \right]_{0}^{1} \int_{0}^{\pi} \ sin(\phi) \ dp d\phi[/tex]

    [tex]volume\ = \frac{2\pi}{3} \int_{0}^{\pi} \ sin(\phi) \ dp d\phi[/tex]

    [tex]volume\ = \frac{4\pi}{3}[/tex]

    Hope this helps

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