How can I find $\Gamma\cap \Omega $ if $\Gamma$ and $\Omega $ are parametrized surface? Let consider two surfaces $$\Gamma=\{\gamma (u,v)\mid u,v\in \mathbb R\}$$
and $$\Omega =\{\omega (u,v)\mid u,v\in \mathbb R\}.$$

How can I find $\Gamma\cap \Omega $ if $\Gamma$ and $\Omega $ are parametrized surface ? Do I have to pass by cartesian coordinates or I can do it directly ? Let take an example $$\Gamma=\{\gamma (u,v)=(\cos u\sin v,\sin u \sin v, \cos v)\mid v\in [0,\pi], u\in [0,2\pi]\}$$
and $$\Omega =\{\omega (u,v)=(u,v,0)\mid u,v\in\mathbb R\}.$$

Let $(x,y,z)\in \Gamma\cap \Omega $. The only thing I can say is that there are $u,v\in[0,2\pi]\times [0,\pi]$ and $s,t\in \mathbb R$ s.t. $$(x,y,z)=\gamma (u,v)=\omega (s,t),$$
but it won't allow me to the circle $$\theta \longmapsto (\cos \theta ,\sin \theta ,0),\quad \theta \in [0,2\pi].$$
What I did is that I find the cartesian equation of $\Gamma$ and $\Omega $ and then parametrized it as following
$$\Gamma: x^2+y^2+z^2=1\quad \text{and}\quad \Omega : z=0,$$
and thus we find $$\Gamma\cap \Omega :\begin{cases}x^2+y^2=1\\ z=0\end{cases},$$
and thus, after parametrization one finds $$\Gamma\cap \Omega =\{c(\theta )=(\cos\theta ,\sin\theta ,0)\mid \theta \in [0,2\pi]\}.$$

Question : First of all, I know that my method using cartesian coordinate is correct, and my question is not to know if the way I solved the problem is correct or not since I know it's correct. My question is : Is there a direct method from 
$$\Gamma=\{\gamma (u,v)\mid u,v\in \mathbb R\}\quad \text{and}\quad \Omega =\{\omega (u,v)\mid u,v\in \mathbb R\},$$
to find $\Gamma\cap \Omega $ (i.e. Without Using Cartesian Coordinate) or I must use them ?

I would thing about something as : finding $u,v$ s.t. $\gamma (u,v)=\omega (u,v)$ but this doesn't work. Other idea would be to fine $u,v,s,t$ s.t. $\gamma (u,v)=\omega (s,t)$, but still doesn't work, so How can I do it if it's possible ?

So I recall the question : How would you find $\Gamma\cap \Omega $ without using coordinate coordinate ?
Calling $\gamma(u,v)=\Gamma$ and $\omega(u,v)=\Omega$ the two intersecting surfaces we have

$$
p_1 = p_1(u_1,v_1) \leftarrow \gamma(u_1,v_1) = \Gamma\\
p_2 = p_2(u_2,v_2) \leftarrow \omega(u_2,v_2) = \Omega
$$

where $p_i = (x_i,y_i,z_i), \;\; i \in \{1,2\}$

For generic surfaces, the intersection can be obtained by solving by an iterative procedure like Newton-Raphson methods. 

So our problem is to solve

$$
p_1(u_1,v_1)-p_2(u_2,v_2) = 0
$$

which gives three equations and four unknowns. The missing equation is used as initial point locator and then as a step constraint. 

This problem can also be handled as a minimization problem or as

$$
\min_{u_1,v_1,u_2,v_2}||p_1-p_2||
$$

I hope to be helpful.
Attached the result for the Newton-Raphson iterative process application to the intersection curve determination. The algorithm used is a simpler one without step uniformity correction.
Calling $\gamma(u,v)=\Gamma$ and $\omega(u,v)=\Omega$ the two intersecting surfaces we have

$$
p_1 = p_1(u_1,v_1) \leftarrow \gamma(u_1,v_1) = \Gamma\\
p_2 = p_2(u_2,v_2) \leftarrow \omega(u_2,v_2) = \Omega
$$

where $p_i = (x_i,y_i,z_i), \;\; i \in \{1,2\}$

For generic surfaces, the intersection can be obtained by solving by an iterative procedure like Newton-Raphson methods. 

So our problem is to solve

$$
p_1(u_1,v_1)-p_2(u_2,v_2) = 0
$$

which gives three equations and four unknowns. The missing equation is used as initial point locator and then as a step constraint. 

This problem can also be handled as a minimization problem or as

$$
\min_{u_1,v_1,u_2,v_2}||p_1-p_2||
$$

I hope to be helpful.
Attached the result for the Newton-Raphson iterative process application to the intersection curve determination. The algorithm used is a simpler one without step uniformity correction.