I am given the two surfaces $$S_1(s,r) = \left( 2\sin r \cdot (e^{s/2}-1)+r, s, e^{s/2}\cdot\sin r \right )$$ and $$S_2(s,r) = (r, s+r, 0)$$ and am asked to find the intersection $S_1\cap S_2$.
I can see that whatever the intersection is, its third coordinate must be $0$. So, $r$ and $s$ must satisfy $$e^{s/2}\cdot\sin r = 0 \implies r=k\pi$$ But, I can't understand how to proceed anymore from here.
In general, I would like to know how to calculate the intersection of the surfaces $$S_1 (r,s) = \left (f_1(r,s), g_1(r,s), h_1(r,s)\right)$$ and $$S_2 (r,s) = \left (f_2(r,s), g_2(r,s), h_2(r,s)\right)$$ where $f_i,g_i,h_i$ are continuous functions.
There is no general method apart from equating each coordinate. In the present case, the most "restrictive" equation is provided by the second coordinate, namely $s = s+r$, hence $r = 0$. In consequence, the first and third equations are satisfied straighforwardly, since they vanish automatically when $r = 0$. Thus, the parameter $s$ remains free and the intersection domain is given by $$ S_1 \cap S_2 = \{(0,s,0) \in \mathbb{R}^3 \;|\; s \in \mathbb{R}\}, $$ which is nothing else than the "$y$-axis".