Surface Integral - Intersection of cone and circle I am now looking at the following exercise:
Calculate: 
$\int \int_S (2z^2 - x^2 - y^2) dS $ where $S $ is $ z=\sqrt{x^2+y^2} $ intersected with $x^2 + y^2 =2x$  (i.e. $ (x-1)^2 + y^2 =1 $ ) . 
Now, as far as definitions:
$\int \int_S (2z^2 - x^2 - y^2) dS  = \int \int_D f(x(u,v),y(u,v),z(u,v)) ||\phi_u \times \phi_v || dudv $  where $\phi:D \to \mathbb{R}^3 $ is the parameterization of $S$ . 
But, how can I parameterize $S$ ? It is the intersection of the cone and the cylinder, but how can I do it?
Thanks in advance
 A: Let's write
$$
S_1 = \{ (x,y,z) \mid z = \sqrt{x^2 + y^2} \}
\quad\text{and}\quad
S_2 = \{ (x,y,z) \mid (x-1)^2 + y^2 = 1\}.
$$
These are surfaces in $\mathbb R^3$, so their intersection $S$ is likely to be a curve and not a surface. Sketching shows that $S$ is an ellipse.
Since $S_2$ is a cylinder is's easy to parametrise it; global coordinates for it are $(\theta,z)$, where $x = 1 + \cos \theta$ and $y = \sin \theta$, where $0 \leq \theta \leq 2\pi$. Plugging this into the equation for $S_1$, we find that the intersection $S$ is parametrised by
$$
\theta \mapsto (1 + \cos\theta, \sin\theta, \sqrt{2 + 2\cos\theta}),
\quad 0 \leq \theta \leq 2\pi,
$$
but the only thing we're really interested in is that $z^2 = 2 + 2\cos \theta$ on $S$. This gives that the function $f$, restricted to $S$, is
$$
f(\theta) = 2 (2 + 2\cos\theta) - (1 + \cos\theta)^2 - \sin\theta^2
= 2 + 2\cos\theta.
$$
Then your integral is
$$
\int_S f \, d\lambda = \int_0^{2\pi} (2 + 2\cos \theta) d\theta = 4\pi.
$$
A: Your surface is the graph of $z = f(x,y) = \sqrt{x^2+y^2}$.  A standard parametrization is
$$(x,y,z)=\phi(x,y)=(x,y,f(x,y))=\left(x,y,\sqrt{x^2+y^2}\right)$$
The domain $D$ in your problem is the disk $x^2 + y^2 \le 2x$
or $-\sqrt{2x-x^2}\le y \le \sqrt{2x-x^2}$, $-1\le y \le1$.
If you convert to polar coordinates, $D$ can be described by $r \le 2 \cos\theta$, $-{\pi\over2}\le \theta \le {\pi\over2}$.
