Lemma 4.4.2. Using the notation of figures 4.6 and 4.9, we have ∂ $_{∂s}$g ( s, r ) = h$_{2}$ − h$_{1}$ = r sin( θ$_{2}$ ) − r sin( θ$_{1}$ ) . Proof. We have The situation is illustrated in figure 4.9 where we can see that the area being added as we go from s to s + Δ s is the shaded region on the right with height r − h$_{1}$ and, considering first-order terms only, uniform width Δ s so has area ( r − h$_{1}$ )Δ s . Similarly, we are subtracting the area ( r − h$_{2}$ )Δ s on the left. Therefore, we have Figure 4.10: T-like data: we restrict the domain of g ( s, r ) to a fixed radius ˆ r plus any vertical segment from r = 0 to r = ˆ r 4.5 Reconstructing shapes from T-like data ∂ ∂s g ( s, r ) = lim Δ s → 0 Area(Ω ∩ D ( γ ( s + Δ s ) , r )) − Area(Ω ∩ D ( γ ( s ) , r )) Δ s . ∂ ∂s g ( s, r ) = lim Δ s → 0 ( r − h$_{1}$ )Δ s − ( r − h$_{2}$ )Δ s Δ s = h$_{2}$ − h$_{1}$. Distance along curve s → s = L r = ˆ r s = 0 s = ˆ s r = 0 r = diam( γ ) Disk radius r → 102