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Volume of a sphere

  1. Dec 2, 2007 #1
    I was just thinking about how you would calculate the volume of a sphere last night, and tried it thusly, using the logic that a sphere is nothing but a bunch of infinitesimally thin circles with increasing radius from 0 to r, and then from decreasing radius r to 0:

    [tex] \int^{R}_{0} \pi r^{2}dr= \frac{1}{3} \pi R^{3}[/tex]

    This, I thought should be one hemisphere, so I multiplied the result by 2, getting [tex]\frac{2}{3} \pi R^{3}[/tex]. This, of course, is not the volume of a sphere. Why did my method not work?
     
  2. jcsd
  3. Dec 2, 2007 #2
    The surface area of a sphere is 4(pi)r^2
     
  4. Dec 2, 2007 #3

    mathwonk

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    you have the radii growing linearly, so you are getting (twice) the volume of a right circular cone of base radius R and height R.

    it seems you should integrate something like pi(R^2 - y^2) as y grows from 0 to R. and then double it. i.e. the thin disc at height y has radius sqrt(R^2-y^2) by pythagoras.


    by the way euclid also used the word "circle" to refer both to the one dimensional circumference and the 2 dimensional interior of a circle, so the flap about that usage is historically unjustified.
     
  5. Dec 2, 2007 #4
    What you calculated was the area below the function [tex]\pi r^2[/tex] above some axis.

    I don't think you can find the volume of a sphere in under 3 integrals since you need to account for the change in x/y/z.

    The integral will be [tex]\int_0^{2\pi}d\theta \int_0^{\pi}sin\phi d\phi \int_0^R r^2 dr[/tex]

    Although it's a tripple integral since none of the limits of one variable depend on another variable you can treat it as the multiple of 3 integrals.
    [tex]\int_0^{2\pi}d\theta[/tex]
    [tex]\int_0^{\pi}sin\phi d\phi[/tex]
    [tex]\int_0^R r^2 dr[/tex]
    After you multiply the results you should get the volume of the sphere [tex]\frac{4}{3}\pi R^3[/tex]
     
  6. Dec 2, 2007 #5

    dst

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    But clearly the starting integral was giving the area of a quadrant of the sphere?
     
  7. Dec 2, 2007 #6

    arildno

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    Your proper integral should be, for the hemishere:
    [tex]\int_{0}^{R}\pi{r}(z)^{2}dz[/tex]
    where z is the variable dsenoting at which height along the axis you are, while r(z) is the radius of the disk at that height.
     
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