Parametrizing Surfaces and Curves

  • #1

king vitamin

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Homework Statement


Given the surface:
x^2 + y^2 + z^2 = 1 but x + y + z > 1 (actually greater than/equal to)

I'd like to parametrize both this portion of the sphere and I'd like to find a parameterization of the boundary of the surface (that is, the intersection of the above sphere and plane).

The Attempt at a Solution


Parametrizing the surface is killing me. If the portion of the sphere were rotated so that it was pointing towards the z-axis it would be parametrized simply by polar coordinates with theta going from 0 to 2Pi and Phi going from 0 to arcsin(3^(-1/2)), but for my problem I need to compute a certain flux (i would need to transform everything, which isn't even taught in this course). The surface isn't a function of any of the coordinate axes, and spherical parameters are useless.

The curve is also lost to me, I'm not sure how to satisfy both x^2 + y^2 + z^2 =1 and x + y + z = 1 with equations involving just one parameter.
 

Answers and Replies

  • #2
My two cents:

I would parametrize it in polar coordinates. I'm not so sure if this is right but I'll give it a whirl.

Let:
x = rcos[tex]\Theta[/tex]
y = rsin[tex]\Theta[/tex]
z = [tex]\sqrt{1-r^2}[/tex]

Thus we can write the parametrization as:

r(r,[tex]\Theta[/tex]) = (rcos[tex]\Theta[/tex])i + (rsin[tex]\Theta[/tex])j + ([tex]\sqrt{1-r^2}[/tex])k

I hope this helps!
 

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