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Parametric Surfaces: rectangular and polar coordinates

  1. Jul 31, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm not grasping how to convert a surface with known rectangular graph to a parametric surface (using some polar techniques, I assume). I would appreciate it if someone could clarify the conversion process.

    One of the examples is as follows:
    A sphere [itex]x^{2}+y^{2}+z^{2}=a^{2}[/itex] is parametrized by [itex]\sqrt{a^{2}-u^{2}}cos(v)\hat{i}+\sqrt{a^{2}-u^{2}}sin{v}\hat{j}+u\hat{k}[/itex]

    2. Relevant equations


    3. The attempt at a solution

    I tried converting the terms using the spherical coordinates: [itex]sin^{2}(\phi)cos^{2}(\theta)+sin^{2}(\phi)sin^{2}(\theta) + cos^{2}(\phi)=a[/itex]
    Last edited: Jul 31, 2012
  2. jcsd
  3. Jul 31, 2012 #2


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    It's not always easy to come up with a parametric equation for some particular object.

    Hopefully you know that for the vector representation of a surface, the vector s a position vector, that is to say, the tail of the vector sits at the origin, while the head of the vector traces out the surface, as the parameter(s) run through their range of values.

    In the case of [itex]\vec{r}=\sqrt{a^{2}-u^{2}}\cos(v)\hat{i}+\sqrt{a^{2}-u^{2}}\sin{v}\hat{j}+u\hat{k}[/itex], we're saying that



    To see that this is a representation of a sphere of radius, a, centered at the origin, square x, y, and z, then take the sum of those squares.

    Of course, we must allow u and v to run though the appropriate set of values.
  4. Jul 31, 2012 #3
    Thanks, that makes sense. It all works so much easier going from parametric to rectangular, but the other way around seems a little far-fetched.
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