# projection R on xy-plane

• Nov 22nd 2007, 02:38 AM
kittycat
projection 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.
• Nov 22nd 2007, 04:16 AM
Manu
projection: ___ is projected on ___
Quote:

Originally Posted by kittycat
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.

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.

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• Nov 22nd 2007, 04:26 AM
Manu
Quote:

Originally Posted by Manu
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...

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 $\displaystyle cos(t)/sin(t) = xy = x(1-x) = x-x^2$
solving for x with the formula $\displaystyle \frac{-b+-sqrt{b^2-4ac}}{2a}$
we get x = $\displaystyle 0.5(1+\sqrt{1+4cos(t)/sin(t)})$

$\displaystyle radius(A) = \sqrt{x^2+y^2} = \sqrt{x^2+(1-x)^2} = \sqrt{2x^2-2x+1}$
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.

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