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Volume of a sphere (solid of revolution)

  1. Jan 25, 2009 #1
    Hi guys, I did post this over on the h/w c/w forum last week, but I think actually it is out of the scope of that, plus its not actually h/w c/w, its my general wondering :D

    So I am having an issue with the volume of a sphere and calculating it. I am using the method where the solid, sphere in this case, is composed of many cylinders, in my case along the x-axis where the center of the sphere is at the origin.

    So initially I attempted this by considering each cylinder was of volume:[itex]\pi y^2 \delta x [/itex]
    then the formula of the circle in the xy plane of the sphere is [itex]x^2 + y^2 = r^2[/itex] which yields [itex]y^2 = {r^2 - x^2} [/itex] which means the volume of each cylinder becomes [itex]\pi (r^2 - x^2) \delta x[/itex]

    I will limit my range to [0,r] as it is symmetrical about the yz plane. So integrating accros the interval [0,r] we get :

    [tex]V = \pi \int_{0}^{r} (r^2 - x^2) \delta x [/tex]
    [tex] = \pi ({r^2}x - \frac{x^3}{3})\bigg{|}_0^r [/tex]
    [tex] = \pi [({r^3} - \frac{r^3}{3})-({r^3}(0) - \frac{0^3}{3})] [/tex]
    [tex] = \pi \frac{2r^3}{3}[/tex]

    then x2 to get the volume of a sphere to be [itex]{4\pi r^3}{/}{3}[/itex]

    now thats all well and good, but I also wanted to see if another method would work, instead of using x as our independent use theta. Now I learnt that if you wanted to calculate the surface are of a sphere you can use theta as it makes calculations easier. So using the idea that we can calculate the surface are of a sphere modeling it as composed of many hollow and cylinders with no base or head, each has height [itex]r\delta \theta[/itex]. So each cylindrical surface becomes :

    [tex]i. \ A = 2\pi y r \delta \theta[/tex]
    [tex]ii. \ y=rsin\theta[/tex]
    [tex]iii. \ A = 2r\pi rsin\theta \delta \theta[/tex]

    So thats fine, that idea can be taken as can successfully yield the surface area of a sphere successfully. Now I figured if that method can be employed for a surface, it must be adaptable to find the volume of a sphere. so this time I decided to define each solid cylinder as:

    [tex]iv. \ V = \pi y^2 r \delta \theta[/tex]
    [tex]v. \ V = r\pi (r^2sin^2\theta) \delta \theta [/tex]

    so again I approached this as before, this time integrating across the range [0, [itex]\pi / 2[/itex]] where I again will only find the hemisphere for +ve x and theta is the angle from the x axis. So my integration went like this :

    [tex]V = \pi r^3\int_{0}^{\frac{\pi}{2}} (sin^2\theta) \delta \theta [/tex]
    [tex] = \pi r^3\int_{0}^{\frac{\pi}{2}} \frac{1}{2}(1 - cos2\theta) \delta \theta [/tex]
    [tex] = \frac{\pi r^3}{2}(\theta - \frac{1}{2}sin2\theta)\bigg{|}_{0}^{\frac{\pi}{2}}[/tex]
    [tex] = \frac{\pi r^3}{2}(\frac{\pi}{2}) = \frac{\pi^2 r^3}{4} [/tex]

    then oviously x2, but as is evident it is not the right answer, I really cant see what I have done wrong, I pretty sure none of my actually mathematics is wrong, which means the only part that could be wrong is the [itex]r\delta \theta[/itex] bit and teacher suggested that for some reason it doesn't apply to a solid but does to a shell, but couldn't give a valid reason as to why. I am inclined to agree with him, but again I cant see why. In various texts I have looked though it says things along the lines of for the shell it is tempting to use [itex]\delta x[/itex] but that [itex]r\delta \theta[/itex] is a "better approximation", therefore surely it to is a better approximation for a solid of revolution to.

    Can anyone explain why this is, what is the reason the integration cant be done this way, or am have I simply done my maths wrong, thanks guys :D
  2. jcsd
  3. Jan 25, 2009 #2


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    If I let [itex]\theta[/itex] be the azimuthal component then I get for the height of the cylinder [itex]r \cos\theta[/itex] with [itex]\theta[/itex] running from 0 to pi.
    Also you need to include the Jacobian from the coordinate transformation into polar coordinates, which I think (since r is fixed) is just sin(theta) - you should check this.
    Then I find
    [tex]\pi r^3 \int_{0}^{\pi} \cos^2\theta (\sin\theta \delta \theta) = \frac{2 \pi r^3}{3} [/tex]
    for half the volume.

    I hope that gets you started.
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