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Linear algebra and volume of rotational object

  1. Dec 1, 2007 #1
    [SOLVED] Linear algebra and volume of rotational object

    1. The problem statement, all variables and given/known data
    I have to show that V(O) = pi * int[f(z)^2]dx when O = {(x,y,z) E R^3 | a =< z =< b, sqrt(x^2+y^2) =< f(z)}.

    I have to integrate 3 times with different limits:

    V(O) = int [dz, a..b] * int[dr, 0 .. f(z)] * int[dTheta * r, 0..2pi].

    Why is it the integral looks like this? I believe it should look like

    int[ int [ int [ dr * dTheta * ds]]] with limits as above.
     
  2. jcsd
  3. Dec 1, 2007 #2

    HallsofIvy

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    You have a< z< b, with a and b constants, so the "outer integral" would be with respect to z, from a to b, in order to cover the entire figure. For each z, you have [itex]\sqrt{x^2+ y^2}< f(z)[/itex]. Taking [itex]r= \sqrt{x^2+ y^2}[/itex], that is the same as r< f(z). Since r (in polar coordinates) cannot be less than 0, the limits of integration with respect to r must be from 0 to f(z) (and the differential is rdr). Finally, there is no limit on [itex]\theta[/itex] so the limits on [itex]\theta[/itex] must be from 0 to [itex]\pi[/itex], the entire circle. That volume is given by
    [tex]\int_{z=a}^{b} \int_{r= 0}^{f(x)} \int_{\theta= 0}^{2\pi} f(z) (d\theta)(rdr)(dz)[/tex]
    It should be easy to see that gives the result you want.

    As for the formula you give, I see no difference except that you have replaced "z" with "s"- without saying what "s" is.
     
  4. Dec 1, 2007 #3
    Why is r^2 = x^2+y^2? And isn't this cylindrical coordinates?

    Thanks for your help so far!
     
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