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Triple integrals

  1. Mar 24, 2005 #1
    Find the mass of a ball B given by "x^2+y^2+z^2≤a^2" if the density at any point is proportional to its distance from the z-axis using cylindrical coordinates So is the density equal to K*sqrt(x^2+y^2), or K*r?


    Using triple integral of f(rcosθ, rsinθ, z)*r*dz*dr*dθ) I got the following: the triple integral of K*r^2dz*dr*dθ, and the limits for the integral w/ respect to z being from -sqrt(a^2-x^2-y^2) to sqrt(a^2-x^2-y^2), which becomes -sqrt(a^2-r^2) to sqrt(a^2-r^2), limits w/ respect to r being from 0 to a and w/ respect to theta from 0 to 2*pi.

    Doing it I find the integral very hard to integrate because I can't do u-subsititon with two r^2's.

    Am I doing anythign wrong? Thanks in advance.
     
    Last edited: Mar 24, 2005
  2. jcsd
  3. Mar 24, 2005 #2

    dextercioby

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    U can very well use spherical coordinates.I'm sure the integration is immediate.

    Daniel.
     
  4. Mar 24, 2005 #3
    I can't. the stupid book asks us to use cylindrical coordinates.

    Is my setup right though?
     
  5. Mar 24, 2005 #4

    dextercioby

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    Why?What's the point...?Why is it hard to integrate...

    [tex] M=k\int_{0}^{2\pi}d\varphi \int_{0}^{a}r^{2} \ dr \int_{-\sqrt{a^{2}-r^{2}}}^{+\sqrt{a^{2}-r^{2}}} \ dz [/tex]

    I think it's trivial.

    Daniel.
     
    Last edited: Mar 24, 2005
  6. Mar 24, 2005 #5
    You forgot a r...in cylindrical coordinates you there's an extra r.
     
  7. Mar 24, 2005 #6

    dextercioby

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    You're right,i edited.Then do it...It's not difficult.You can make a "sin/cos" substitution and solve it easily.

    Daniel.
     
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