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Surface integral

  1. Mar 23, 2009 #1
    I want to integrate something in spherical coordinates
    I have [tex] da=R^{2}sin(g)dgdh \hat{r} [/tex] with g and h angles
    and [tex] \hat{r}=sin(g)cos(h) \hat{i}+sin(g)sin(h) \hat{j}+cos(g) \hat{k} [/tex]

    But what is now [tex] da_{x}=dydz \hat{i} [/tex] in spherical coordinates?
    So I have the expression in ordinary coordinates and need to find it in spherical coordinates

  2. jcsd
  3. Mar 23, 2009 #2


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    On a spherical surface of radius 1, [itex]\vec{r}= sin(\phi)cos(\theta)\vec{i}+ sin\phi)sin(\theta)\vec{j}+ cos(\phi)\vec{k}[/itex] as you say.

    Differentiating with respect to each variable,
    [itex]\vec{r}_\theta= -sin(\phi)cos(\theta)\vec{i}+ sin(\phi)cos(\theta)\vex{j}[/itex]
    [itex]\vec{r}_\phi= cos(\phi)cos(\theta)\vec{i}+ cos(\phi)sin(\theta)\vec{j}- sin(\phi)\vec{k}[/itex]

    The "fundamental vector product" of a surface is the cross product of those two derivative vectors:
    [tex]\left|\begin{array}\vec{i} & \vec{j} & \vec{k} \\ -sin(\phi)sin(\theta) & sin(\phi)cos(\theta) & 0 \\ cos(\phi)cos(\theta) & cos(\phi)sin(\theta) & -sin(\phi)\end{array}\right|= -sin^2(\phi)cos(\theta)\vec{i}- sin^2(\phi)sin(\theta)\vec{j}- sin(\phi)cos(\phi)\vec{k}[/tex]
    has length [itex]sin(\phi)[/itex] so the differential of surface area is [itex]sin(\phi)d\phi d\theta[/itex].

    (In general if a surface is given by [itex]\vec{r}(u, v)[/itex] with parameters u and v, then the differential of surface area is [itex]\left|\vec{r}_u\times\vec{r}_v|dudv[/itex]. Thats's worth knowing! In fact, the "vector differential" [itex]\vec{r}_u\times\vec{r}_v du dv[/itex] is a vector having the differential of surface area as length, normal to the surface and can be used to integrate vector fields over the surface.)
    Last edited by a moderator: Mar 23, 2009
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