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Math Help - Cartesian -> Spherical polar

  1. #1
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    Cartesian -> Spherical polar

    I have an integral \int \int_S  x^2 + yz \ dS

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

     dS = r^2 \sin \theta d\theta d\phi ??

    Where surface S is x^2 + y^2 + z^2 = 1
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  2. #2
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    Maybe this will help:

    Spherical coordinate system - Wikipedia, the free encyclopedia

    Scroll down to where it says:"Integration and differentiation in spherical coordinates"
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    I've seen that page, but how is dS = r^2 \sin \theta d\theta d\phi arrived at?
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    Member eXist's Avatar
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    Ok I think I got it, but lets see if you can follow me here. You know how polar coordinates work right?

    Where: ds = rdrd\theta
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    Quote Originally Posted by eXist View Post
    Ok I think I got it, but lets see if you can follow me here. You know how polar coordinates work right?

    Where: ds = rdrd\theta
    I'm afraid I don't follow that...
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  6. #6
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    eXist is referring to two dimensional polar coordinates. I admit I don't see what that has to do with your question!

    You asked this same question on "Physics Forums" and I answered it there- but here goes again:

    There are two ways of looking at this.

    First, a rather rough "geometrical" look. Imagine an infinitesmal "rectangle" on the sphere having "top" at \phi and "bottom" at \phi+ d\phi, "left" at \theta and "right" at \theta+ d\theta. 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 1(d\phi)= d\phi. 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 \phi. Then r/1= sin(\phi) so the radius of the circle is sin(\phi) and the length of the arc on the sphere is sin(\phi)d\theta. The area of the "rectangle" is (width times height) sin(\phi)d\theta d\phi.

    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: \vec{r}= f(u,v)\vec{i}+ g(u,v)\vec{j}+ h(u,v)\vec{k}. The derivatives of that with respect to the two variables, \vec{r_u}= f_u(u,v)\vec{i}+ g_u(u,v)\vec{j}+ h_u(u,v)\vec{k} and \vec{r_v}= f_v(u,v)\vec{i}+ g_v(u,v)\vec{j}+ h_v(u,v)\vec{k} are vectors in the tangent plane to the surface at each point. The cross product then, \vec{r_u}\times\vec{r_v}, the "fundamental vector product", is perpendicular to the surface and, because of the derivatives, its length measures the area of an infinitesmal region: dS= \left|\vec{r_u}\times\vec{r_v}\right|dudv.

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

    By the way, my \theta and \phi are reversed from yours. Mathematics notation (mine) uses \theta as "longitude" and \phi as "co-latitude". Engineering notation reverses those.
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    Thanks for taking the time to write that explanation, it helps to clarify things!

    For anyone who finds this post later, there are a couple of nice diagrams which illustrate HallsofIvy's post:
    http://www.netcomuk.co.uk/~jenolive/polar.html
    Last edited by bigdoggy; August 26th 2009 at 02:33 PM.
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    Quote Originally Posted by HallsofIvy View Post
    By the way, my \theta and \phi are reversed from yours. Mathematics notation (mine) uses \theta as "longitude" and \phi as "co-latitude". Engineering notation reverses those.
    And physicists and mathematicians in the UK. I'm not really sure why the difference in convensions!
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