I have an integral
and wish to transform to spherical polar coordinates. How does dS become
Where surface S is
Maybe this will help:
Spherical coordinate system - Wikipedia, the free encyclopedia
Scroll down to where it says:"Integration and differentiation in spherical coordinates"
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 and "bottom" at , "left" at and "right" at . 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 . 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 . Then so the radius of the circle is and the length of the arc on the sphere is . The area of the "rectangle" is (width times height) .
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: . The derivatives of that with respect to the two variables, and are vectors in the tangent plane to the surface at each point. The cross product then, , the "fundamental vector product", is perpendicular to the surface and, because of the derivatives, its length measures the area of an infinitesmal region: .
The spherical coordinates are connected to the cartesian coordinates by , , and we can take those as parametric coordinates for the unit sphere by taking : , , or . The derivatives are and . The cross product of those is which has length . That gives .
By the way, my and are reversed from yours. Mathematics notation (mine) uses as "longitude" and as "co-latitude". Engineering notation reverses those.
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: