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Cartesian -> Spherical polar

  1. Aug 26, 2009 #1
    I have an integral [tex]\int \int_S x^2 + yz \ dS[/tex]

    and wish to transform to spherical polar coordinates. How does dS become

    [tex] dS = r^2 \sin \theta d\theta d\phi [/tex]??

    Where surface S is [tex]x^2 + y^2 + z^2 = 1[/tex]
     
  2. jcsd
  3. Aug 26, 2009 #2

    HallsofIvy

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    There are two ways of looking at this.

    First, a rather rough "geometrical" look. Imagine an infinitesmal "rectangle" on the sphere having "top" at [itex]\phi[/itex] and "bottom" at [itex]\phi+ d\phi[/itex], "left" at [itex]\theta[/itex] and "right" at [itex]\theta+ d\theta[/itex]. The left and right, "lines of longitude", are great circles, with radius the radius of the sphere, 1. The length of those two sides is [itex]1(d\phi)= d\phi[/itex]. The top and bottom, "lines of latitude", are not great circles. Their centers lie on the line through the poles, (0,0,1) and (0,0,-1). Given a point on one of those circles, drop a perpendicular to that vertical line. You have a right triangle with one leg being the radius of the circle, r, the hypotenuse the radius of the circle, 1, and angle at the center [itex]\phi[/itex]. Then [itex]r/1= sin(\phi)[/itex] so the radius of the circle is [itex]sin(\phi)[/itex] and the length of the arc on the sphere is [itex]sin(\phi)d\theta[/itex]. The area of the "rectangle" is (width times height) [itex]sin(\phi)d\theta d\phi[/itex].

    A more rigorous calculation involves the "fundamental vector product" which is worth knowing on its own. Any smooth surface, since a surface is two dimensional, can be written in terms of two parameters: x= f(u,v), y= g(u,v), z= h(u,v) which can also be written as the vector equation: [itex]\vec{r}= f(u,v)\vec{i}+ g(u,v)\vec{j}+ h(u,v)\vec{k}[/itex]. The derivatives of that with respect to the two variables, [itex]\vec{r_u}= f_u(u,v)\vec{i}+ g_u(u,v)\vec{j}+ h_u(u,v)\vec{k}[/itex] and [itex]\vec{r_v}= f_v(u,v)\vec{i}+ g_v(u,v)\vec{j}+ h_v(u,v)\vec{k}[/itex] are vectors in the tangent plane to the surface at each point. The cross product then, [itex]\vec{r_u}\times\vec{r_v}[/itex], the "fundamental vector product", is perpendicular to the surface and, because of the derivatives, its length measures the area of an infinitesmal region: [itex]dS= \left|\vec{r_u}\times\vec{r_v}\right|dudv[/itex].

    The spherical coordinates are connected to the cartesian coordinates by [itex]x= \rho cos(\theta)sin(\phi)[/itex], [itex]y= \rho sin(\theta)sin(\phi)[/itex], [itex]z= \rho cos(\phi)[/itex] and we can take those as parametric coordinates for the unit sphere by taking [itex]\rho= 1[/itex]: [itex]x= cos(\theta)sin(\phi)[/itex], [itex]y= sin(\theta)sin(\phi)[/itex], [itex]z= cos(\phi)[/itex] or [itex]\vec{r}= cos(\theta)sin(\phi)\vec{i}+ sin(\theta)sin(\phi)\vec{j}+ cos(\phi)\vec{k}[/itex]. The derivatives are [itex]\vec{r_\theta}= -sin(\theta)sin(\phi)\vec{i}+ cos(\theta)sin(\phi)\vec{j}[/itex] and [itex]\vec{r_\phi}= cos(\theta)cos(\phi)\vec{i}+ sin(\theta)cos(\phi)\vec{j}- sin(\phi)\vec{k}[/itex]. The cross product of those is [itex]cos(\theta)sin^2(\phi)\vec{i}+ sin(\theta)sin^2(\phi)\vec{j}+ sin(\phi)cos(\phi)\vec{k}[/itex] which has length [itex]\sqrt{cos^2(\theta)sin^4(\phi)+ sin^2(\theta)sin^4(\phi)+ sin^2(\phi)cos^2(\phi)}[/itex][itex]= \sqrt{sin^4(\phi)+ sin^2(\phi)cos^2(\phi)}[/itex][itex]= \sqrt{sin^2(\phi)(sin^2(\phi)+ cos^2(\phi))}[/itex][itex]= sin(\phi)[/itex]. That gives [itex]dS= sin(\phi)d\phi d\theta[/itex].

    By the way, my [itex]\theta[/itex] and [itex]\phi[/itex] are reversed from yours. Mathematics notation (mine) uses [itex]\theta[/itex] as "longitude" and [itex]\phi[/itex] as "co-latitude". Engineering notation reverses those.
     
    Last edited: Aug 26, 2009
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