Surface integral: Cone cut by a cylinder Ok I've got this exercise from Apostol I'm trying to do:
"The cylinder $x²+y²=2x$ cuts out a portion of a surface S from the upper nappe of the cone x²+y²=z². Compute the value of the integral:
$$\int\int_S(x^4-y^4+y^2z^2-z^2x^2+1)dS$$
Ok, what I've done so far is choosing a parametrization $X(u,v)=vcosu$    $Y(u,v)=vsinu$ and $Z(u,v)=v$
I'm not sure the parametrization is good but It really seems it is.
I've seem some really close examples solved at other topics but my question is not about exactly the same thing. I know what I have to do mostly but I can't figure out how am I supposed to set the restrictions on the parameters u and v so that they just run thru the surface wich lies inside de cylinder, in other words, how am I supposed to set the boundings of integration? This is my problem in every exercise....
(extra)Also I'd like to know how could I use the Theorem of Implicit Functions to find the normal vector of intersection of surfaces in $R^3$.

The points $(x,y,z)$ on the surface satisfy $x^2+y^2 \le 2x$, $z^2 = x^2+y^2$, $z = 0$. 
Since you set $X(u,v) = v\cos u$, $Y(u,v) = v\sin u$, and $Z(u,v) = v$ we get: 
$z \ge 0 \leadsto v \ge 0$
$x^2+y^2 \le 2x \leadsto (v\cos u)^2+(v\sin u)^2 \le 2(v\cos u) \leadsto v^2 \le 2v\cos u \leadsto 0\le v \le 2\cos u$. 
In order for $0 \le 2\cos u$, we need $\cos u \ge 0$, i.e. $-\dfrac{\pi}{2} \le u \le \dfrac{\pi}{2}$. 
Therefore, the bounds of $u,v$ are $-\dfrac{\pi}{2} \le u \le \dfrac{\pi}{2}$ and $0 \le v \le 2\cos u$.

Since the surface is parameterized by $\vec{r} = (v \cos u)\hat{i} + (v\sin u)\hat{j} + v\hat{k}$, we can compute the fundamental vector product to get a normal vector to the surface as follows: 
$\vec{N}(u,v) = \vec{r}_u \times \vec{r}_v$ $= \left|\begin{matrix}\hat{i}&\hat{j}&\hat{k}\\-v\sin u & v\cos u & 0\\ \cos u & \sin u & 1\end{matrix}\right|$ $= (v\cos u)\hat{i}+(v\sin u)\hat{j}-v\hat{k}$.
