I am trying to determine the surface area of a torus that has been twisted. I know the non-twisted case can be solved by either using Pappus's Theorem or a method along the lines of this post.
To be more precise, suppose we take a circle of radius $1$ sitting in the $xz$-plane with center at $(2,0,0)$ and revolve it about the $z$-axis so that the center of the revolving circle always rests in the $xy$-plane.
Right before attaching this rotated object back up upon itself to form a torus, we rotate the circle by some amount $\theta$ and then attach the starting circle to the rotating circle.
From the outside, this torus looks just like a normal torus, but it certainly is not because if we consider where the point $(3,0,0)$ goes in the process of the revolution, it is clear that some nonzero rotation will increase the length of the path sweeped out by the point.
My question is, how (if?) does the area change? Moreso, how do you calculate this new area? (I think this may depend upon the "fervor" in which which one twists the torus. For example, if we hope to acheive a twist by $\pi/2$, we can do it in the last $25\%$ of the revolution or the last $50\%$ of the revolution (actually, I have no idea if the area would change in this case!).