Torus Equation

    • Thread Starter

    Let T be a circle with diameter AC=a, and let B be a point on T such that AB=b<a. Let T' be the semicircle with diameter AC in the plane perpendicular to that of T.

    Rotating T' around the perpendicular at A to the plane ABC generates a torus  \mathcal{T} of internal radius 0. Give the equation of  \mathcal{T} .

    I've started by drawing everything out. If the circle is in the x-y plane, with the point A as at the origin, then the perpendicular is the z axis and the semi circle T' is in the x-z plane. Rotating around the z axis I can't see how this produces a torus? At best I get an upper half of a torus?

    (Original post by miml)
    At best I get an upper half of a torus?
    Yep, that's what I make it.

    Edit: There also seems to be redundant information there. Multipart question?
    • Thread Starter

    (Original post by ghostwalker)
    Yep, that's what I make it.

    Edit: There also seems to be redundant information there. Multipart question?
    Thanks, I think what they want is to extend the semi-circle to a circle in the x-z plane, rotate it, give the equation of the resulting torus and then for the following parts only consider the upper half.

    Archytas' solution to doubling the cube if you're interested.
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