Surface S is the portion of the plane x+y+z=1 IN THE FIRST OCTANT, oriented by unit normals with positive components.

Is the projection R = { (r, theta) such that 0< or = r < or = 1, 0 < or = theta < or = pi/2} ?

Thank you very much.

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- November 22nd 2007, 02:38 AMkittycatprojection R on xy-plane
Surface S is the portion of the plane x+y+z=1 IN THE FIRST OCTANT, oriented by unit normals with positive components.

Is the projection R = { (r, theta) such that 0< or = r < or = 1, 0 < or = theta < or = pi/2} ?

Thank you very much. - November 22nd 2007, 04:16 AMManuprojection: ___ is projected on ___
I think your question is not complete, are you projecting the surface S on the XY plane? are you projecting something on the surface S? projection is meaningless unless you say what on what...

Anyway,

The X+Y+Z=1 in the first octant looks like the face of a pyramid,

when projected on the XY plane it will look like a filled triangle.

The R looks like a quarter of a ball, so they aren't the same.

To solve the (r,t) surface of the triangle is not so simple, so I won't do it unless you say this is what you want.

**Yuwie.com**| invite friends. hang out. get paid. - November 22nd 2007, 04:26 AMManu
Ooops, I just saw in your title that you did describe more details,

so if I understand correctly you need to project the S surface on the XY plane and describe it in (r,t) surface?

if this is the case, the simplest would be to go with t from 0 to pi/2, and then with r from 0 to radius(A),

where A is the point that has the same ratio as (cos(t),sin(t)) but its components when added together are one...

so A is (x,y) such that x+y=1 and

solving for x with the formula

we get x =

put in the x we got and you get the radius(A) by the angle t,

and then you can write the (r,t) shape... gag it looks bad, but there is just no pretty way to write x+y=1 in (r,t) form.

**Yuwie.com**| invite friends. hang out. get paid.