Parametric Surfaces: rectangular and polar coordinates

[hsetennis]

Homework Statement

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]

Homework Equations

None.

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]

 

[SammyS]

Homework Statement

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]

Homework Equations

None.

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]

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

[itex]x=\sqrt{a^{2}-u^{2}}\cos(v)[/itex]

[itex]y=\sqrt{a^{2}-u^{2}}\sin(v)[/itex]

[itex]z=u[/itex]​

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.

 

[hsetennis]

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.