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Math Help - projection R on xy-plane

  1. #1
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    Question 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.
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  2. #2
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    projection: ___ is projected on ___

    Quote Originally Posted by kittycat View Post
    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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  3. #3
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    Quote Originally Posted by Manu View Post
    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 cos(t)/sin(t) = xy = x(1-x) = x-x^2
    solving for x with the formula \frac{-b+-sqrt{b^2-4ac}}{2a}
    we get x = 0.5(1+\sqrt{1+4cos(t)/sin(t)})

     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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