Volume form of a hyperbolic knot Given a hyperbolic knot $K: S^1 \to S^3 := \mathbb{R}^3 \cup \{\infty\}$, there is a unique metric $g$ on the complement of the image with constant negative curvature, by Mostow rigidity theorem.
Assume that $K$ doesn't touch $\infty$. Can the volume form $dV_g$ be described from the three functions
$$K_i(\theta)\qquad\qquad\qquad \theta \in [0,2\pi),\quad i = 1, 2, \text{or 
 } 3?$$
 A: The Mostow Rigidity Theorem is not quite stated quite correctly in your post: $g$ is not unique on the nose as you have stated it. Instead, the correct statement is that if $g,g'$ are two complete, finite volume Riemannian metrics on $S^3 - K(S^1)$ of constant sectional curvature $-1$ then there exists a diffeomorphism $h : S^3 - K(S^1) \to S^3 - K(S^1)$ homotopic to the identity such that $g = h^*(g')$ (Gabai strenghthened this conclusion to "isotopic to the identity").
Because of this, it is not quite reasonable to ask for the volume form. One could instead ask for a volume form, in fact one could ask for more, one could ask for a Riemannian metric $g$ satisfying the desired conditions (and then with $g$ in your hands you could compute the volume form $dV_g$ at your leisure). But the point to realize here regarding uniqueness is that one can always perturb that metric $g$, by choosing a small diffeomorphism isotopic to the identity, and pushing $g$ forward to get another such metric $g'$.
So, having said all of that, the best answer I can think of in full generality is to use SnapPea to compute a decomposition of $S^3 - K(S^1)$ into ideal tetrahedra, together with a corresponding gluing diagram of ideal $\mathbb H^3$ tetrahedra. One could then write down $g$ for oneself, by using the SnapPea output to map the ideal $\mathbb H^3$ tetrahedra in that gluing diagram to the ideal tetrahedra of the given triangulation, then pushing forward the hyperbolic metrics from the ideal $\mathbb H^3$ tetrahedra.
