Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Formulae derivation

  1. Mar 12, 2005 #1


    User Avatar

    I'm just a little curious with these formulas because I like to understand a formula before I use it. There's no doubt I've been using these since gr3 but I would like a mathmatical derivation of them.

    All I would like to know here is why by 3?

    And just
    [tex]V_{sphere}=\frac{4\pi r^3}{3}[/tex]
    With this one I have no clue why it is what it is. There are [itex]4\pi [/itex]radians in a sphere maybe??

    Thanks in advance.
  2. jcsd
  3. Mar 12, 2005 #2
    hm do you mean where does 3 come from?
    in integration (calculus), we can find the formula by integration.
    for ball, find antiderivative of p f(x)^2 dx
  4. Mar 12, 2005 #3
    Well, a cone of height [tex]h[/tex] and with base radius [tex]\alpha h[/tex] can defined by [tex]T = \{ (x,\ y,\ z) \ | \ (\alpha z)^2 \geq x^2 + y^2,\; 0\leq z \leq h \}[/tex]. Clearly it is a solid of revolution (it is symmetric about the z-axis). Its volume is

    [tex]V_{\mbox{cone}} = \int \int \int_T dV = \int_0^h \int_{-\alpha z}^{\alpha z} \int_{-\sqrt{(\alpha z)^2 - x^2}}^{\sqrt{(\alpha z)^2 - x^2}} \ dy \ dx\ dz = 2 \int_0^h \int_{-\alpha z}^{\alpha z} \sqrt{(\alpha z)^2 - x^2} \ dx \ dz = 2\int_0^h {(\alpha z)^2 \pi \over 2} \ dz = \alpha^2 \pi \left[{z^3\over 3}\right]_0^h = {\alpha^2 \pi h^3 \over 3}[/tex]

    From here all you need to notice is that the base area is [tex] A_b = \alpha^2 h^2 \pi \Longrightarrow V_{\mbox{cone}} = {\alpha^2 \pi h^3 \over 3} = {A_b h \over 3}[/tex] as we wanted. Note that the third-to-last step is valid:

    [tex]\int_{-a}^a \sqrt{a^2-x^2} dx = \int_{-{\pi \over 2}}^{\pi \over 2} a^2 \cos^2{t} \ dt = a^2\int_{-{\pi \over 2}}^{\pi \over 2} {1\over 2}\left[1+\cos{2t}\right] \ dt = {a^2 \over 2}\left[t + {\sin{2t} \over 2}\right]_ {-{\pi \over 2}}^{\pi \over 2} = {a^2 \pi \over 2}[/tex]

    This is, incidentally, the area of a semicircle of radius [tex]a[/tex], which is to be expected since [tex]0 \leq y \leq \sqrt{a^2-x^2}, -a \leq x \leq a[/tex] defines such a semicircle!

    The pyramid can be done in a similar manner, or by appealing to its geometry. You can try it :)

    The volume of a sphere is also found in the same way, although this one's easier because you can just use spherical coordinates (who would have guessed!). A sphere of radius [tex]r[/tex] can be defined in polar coordinates to be just [tex] S = \{ ( \rho, \ \phi, \ \theta ) \ | 0 \leq \rho \leq r, \ 0 \leq \phi \leq \pi, \ 0 \leq \theta < 2\pi \} [/tex]. Using the Jacobian transformation we find that since our coordinate transformation is [tex] x = \rho \sin{\phi} \cos{\theta}, \ y = \rho \sin{\phi} \sin{\theta}, z = \rho \cos{\phi}[/tex] we get [tex] dV = \rho^2 \sin{\phi} \ d\rho \ d\phi \ d\theta [/tex] and so our volume is

    [tex] V_{\mbox{sphere}} = \int \int \int_S dV = \int_0^{2\pi} \int_0^\pi \int_0^r \rho^2 \sin{\phi} \ d\rho \ d\phi \ d\theta = 2\pi \int_0^\pi \sin{\phi} \ d\phi \int_0^r \rho^2 \ d\rho = 2\pi\left[-\cos{\theta}\right]_0^\pi\left[{\rho^3\over 3}\right]_0^r = (2\pi)(2){r^3 \over 3} = {4 \pi r^3 \over 3}[/tex]

    again as we wanted. As to your question about [tex] 4\pi[/tex] radians in a sphere: not quite, but very close. There are [tex]4\pi [/tex] of what we call steradians in a sphere. These are a "unit" (and I use the term loosely, because neither radians nor steradians are really a unit) for a sort of two-dimensional angular arc.
    Last edited: Mar 12, 2005
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook