Year 12s: what will be the first way you research your future options? >>

Torus Equation

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?

Reply 1

ghostwalker

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?

Reply 2

miml

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.