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Homework Help: Derivation FOR THE SURFACE AREA OF A SPHERE

  1. Feb 24, 2009 #1
    [tex] 4 \pi r^2 = SA [/tex] for a sphere. everybody knows that.

    but, how do you derive this by integrating infinitesimal amounts of area over the curvature of the sphere????
     
  2. jcsd
  3. Feb 24, 2009 #2

    lanedance

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    think of a curve you could rotate around an axis to get a sphere

    an surface area element is dA = 2.pi.ds, with ds an infinitesimal length along curve... imagine unrotating a tiny strip
     
  4. Feb 24, 2009 #3

    lurflurf

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    There are many ways
    -integrate over the area
    note r^2=x^2+y^2+z^2
    or use parameters like
    x=cos(s)cos(t)
    y=cos(s)sin(t)
    z=sin(s)
    -project a circle
    the projection will be 4 circles
    4*pi*r^2
    -consider the solid of revolution of a semi circle
    -differentiate the volume
    [(4/3)pi*r^3]'=4pi*r^2
     
  5. Feb 24, 2009 #4

    HallsofIvy

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    The "standard" way of finding surface area (although I like orthovector's method, treating it as a semicircle rotated around the y-axis better):

    In spherical coordinates, [itex]x= \rho cos(\theta)sin(\phi)[/itex], [itex]y= \rho sin(\theta)sin(\phi)[/itex], [itex]z= \rho cos(\phi)[/itex].

    Taking [itex]\rho[/itex] constant, say [itex]\rho= R[/itex], gives the surface of the sphere of radius R in terms of parameters [itex]\theta[/itex] and [itex]\rho[/itex] and we can write the vector equation as
    [tex]\vec{r}= Rcos(\theta)sin(\phi)\vec{i}+ Rsin(\theta)\sin(\phi)\vec{j}+ Rcos(\phi)\vec{k}[/tex]
    Differentiating with respect to [itex]\theta[/itex] and [itex]\phi[/itex] gives to tangent vectors to that surface
    [tex]\vec{r}_\theta= -Rsin(\theta)sin(\phi}\vec{i}+ Rcos(\theta)sin(\phi)\vec{j}[/tex]
    [tex]\vec{r}_\phi= Rcos(\theta)cos(\phi)\vec{i}+ Rcos(\theta)cos(\phi)\vec{j}- Rsin(\phi)\vec{k}[/tex]

    The "fundamental vector product" for the surface is the cross product of those two tangent vectors:
    [tex]R^2cos(\theta)sin^2(\phi)\vec{i}- R^2sin(\theta)sin^2(\phi)\vec{j}- R^2sin(\phi)cos(\phi)\vec{k}[/tex]
    is perpendicular to the surface and its length, [itex]R^2 sin(\phi)[/itex] gives the "differential of surface area", [itex]R^2 sin(\phi) d\theta d\phi[/itex]. To cover the entire surface of the sphere, [itex]\theta[/itex] must range from 0 to [itex]2\pi[/itex] and [itex]\phi[/itex] must range from 0 to [itex]\pi[/itex]. The surface area is given by
    [tex]\int_{\theta= 0}^{2\pi}\int_{\phi= 0}^\pi R^2 sin(\phi)d\phi d\theta[/tex]

    Integrating with respect to [itex]\theta[/itex] immediately gives
    [tex]2\pi R^2 \int_{\phi= 0}^\pi sin(\phi)d\phi= -2\pi R^2 cos(\phi)\right|_0^\pi[/tex]
    [tex]= -2\pi R^2 \left(-1- 1)= 4\pi R^2[/tex]
     
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