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Rotating 2D coordinate geometry/conic sections into 3D - how?

  1. Apr 25, 2007 #1
    I know basic 2D coordinate geometry/conic sections and some vectors and calculus. What Im trying to find out is, if you rotate a conic about its axis by an angle of pi, you should get a surface in 3D. How would you go about rotating this though?
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  3. Apr 25, 2007 #2
    Let's say you have an ellipse lying entirely on the x-y plane, symmetrical about the x- and y- axis. (Your typical ellipse.) When you rotate this plane figure about either of those axes (by 2pi) you end up with an ellipsoid. You have to imagine this happening in 3-D space. For something simpler, make that ellipse into a circle; you will end up with a sphere.
  4. Apr 25, 2007 #3


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    What do you mean by "go about rotating this"? If you draw a parabola in the xz-plane with the postive z-axis as axis of symmetry, and rotate about that axis, you get an "elliptic paraboloid" with equation z= x2+ b2. One way to see that is to take the axis perpendicular to the z-axis to be "r", in polar coordinates. That way, z= r2 becomes z= x2+ y[/sup]. (There is also a "hyperbolic paraboloid" but it is not a "surface of revolution".)

    Similarly, rotating an ellipse, [itex]b^2r^2+ a^2z^2= a^2b^2[/itex] around the z-axis gives an "ellipsoid" [itex]b^2x^2+ b^2y^3+ a^2z^2= a^2b^2[/itex] or, equivalently, [itex]x^2/a^2+ y^2/a^2+ z^2/b^2= 1[/itex]. You have either a "prolate" or "oblate" ellipsoid depending upon whether the axis of rotation is the major or minor axis respectively. The third possibility is [itex]x^2/a^2+ y^2/b^2+ z^2/c^2= 1[/itex] where a, b, and c are all different but, again, that is not a surface of revolution.

    You get distinctly different surfaces if you rotate a hyperbola around its two different axes. Rotating around that passes through the hyperbola, you get a "hyperboloid of two sheets" since the two parts of the hyperbola for two separated surfaces. Now you would have [itex]r^2/a^2- z^2/b^2= x^2/a^2+ y^2/a^2- z^2/b^2= 1[/itex].
    If you rotate around the axis that is does NOT go through the hyperbola you get the "hyperbola of one sheet" since the two parts of the hyperbola rotate through the same surface. You would have [itex]z^2/a^2- r^2/b^2= z^2/a^2- x^2/a^2- y^2/a^2= 1[/itex]. Of course, you could have all the variations on [itex]x^2/a^2- y^2/b^2- z^2/c^2= 1[/itex] with different number and placements of the negative signs but with a, b, c all different those are not surfaces of rotation.

    I'll leave it to you to decide what figure you get, and what its equation is if rotate a circle around an axis!
  5. Apr 26, 2007 #4
    Ok. What I meant was, how do you mathematically derive these equations? Like do you take the equation of a hyperbola in the plane z=0, and then trace the same in a plane rotated by an angle [tex]d\theta[/tex] and then integrate the equation from 0 to pi (you would rotate from 0 to pi because there are two arms of the hyperbola) about the transverse axis. Would this work?

    For a sphere, I guess you could take a point circle at x=0 (in the yz plane), and then another circle with radius r+dr in the plane a distance dx away from the yz plane and then integrate this from 0 to 2r to get the sphere. Now, all this I can imagine and do in my mind, but to put this down on paper, is what Im having trouble doing. I've never seen anything like this before. I know the equation of the sphere would be of the form [tex]x^2+y^2+z^2+2wx+2uy+2vz+c=0[/tex], and given this equation, I could find the center, radius, and maybe even figure out the equation of the tangent at a point. But, I dont know how this equation came to be. Thats what I'm trying to find out.

    Now, I understand I'm trying to trace the surface in space. I dont know how to go about finding the locus of the point I'm trying to trace. The properties of the surface should be valid in all planes, so, if for example, I take the equation of a parabola, and then find the equation of its directrix, find the equation of the plane perpendicular to the axis of the parabola which contains the directrix, then the locus of the point which is always equidistant from the focus of the initial parabola and the plane should be my surface. That would work, right?
    Last edited: Apr 26, 2007
  6. Apr 27, 2007 #5
    No, that wont work!!! How would you trace this surface?
  7. Apr 27, 2007 #6


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    The easiest way to do it is using parametric equations.

    Say you have a curve:
    C(t) = <x(t), y(t)>, a <= t <= b

    Then to rotate it around the x axis:
    S(t) = <x(t), y(t)*cos(theta), y(t)*sin(theta)>, a <= t <= b, 0 <= theta <= 2pi

    And to rotate it around the y axis:
    S(t) = <x(t)*cos(theta), y(t), x(t)*sin(theta)>, a <= t <= b, 0 <= theta <= 2pi
  8. Apr 27, 2007 #7


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    I thought I had just explained that. Take some figure whose equation in the xy-plane would be F(x,y)= 0. (Ellipse, [itex]F(x,y)= x^2/a^2+ y^2/b^2-1=0[/itex]; parabola, [itex]F(x,y)= y- ax^2= 0[/itex]; hyperbola, [itex]F(x,y)= x^2/a^2- y^2/b^2-1= 0[/itex]; etc. but this idea applies to any figure that can be written F(x,y)= 0.) To rotate such a figure around the z-axis, replace y by z and x by r. For an ellipse, for example, you would have [itex]r^2/a^2+ z^2/b^2= 1[/itex]. That gives you the equation in "cylindrical coordinates". To go back to Cartesian coordinates, replace that "r" by [itex]\sqrt(x^2+ y^2)[/itex]. Then you have, for the ellipsoid, [itex]x^2/a^2+ y^2/a^2+ z^2/b^2= 1[/itex]
  9. Apr 27, 2007 #8
    Ok, What you're doing here is taking any arbitrary plane perpendicular to the z axis on which you define circles of varying radii. Those radii are also governed by their position on the z-axis (thats what the term [tex]z^2/b^2[/tex] signifies, right?) and then treating this collection of circles as your surface right? I'm sorry I'm being difficult here. Why did you replace [tex] r^2[/tex] by [tex]x^2+y^2[/tex]? Again, I'm sorry I'm asking you so many questions, but please answer them.
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