Double Integral: Find the flux through a surface. Check my answer The Question:
Let $\Sigma$ be the portion of $z=16-x^2-y^2$ inside the cylinder $r=2cos\theta$ and with upward orientation. Draw a picture of $\Sigma$ and find the rate at which the fluid $\mathbf F = x\mathbf j$ is flowing through $\Sigma$
Please check my solution. I attempted it using an oriented surface integral
The picture is a upside down paraboloid with bounds of a circle with radius $2$ centered at $x=1$
I first parametrized the surface. 
$\mathbf r = x \mathbf i + y \mathbf j + (16-x^2-y^2 )\mathbf k$
Next I found the cross product: $\mathbf r_x \times \mathbf r_y = 2x \mathbf i + 2y \mathbf j + \mathbf k$
The cross product is oriented upward.
And my final integral is:
$\int_{-\pi/2}^{\pi/2}\int_0^{2cos\theta}\ 2r^3cos\theta sin\theta\, drd\theta$
 A: What you did looks right!
Method 1
Parametrize as you did:
$$
x=x, \quad y=y, \quad z=16-x^2-y^2,
$$
with


*

*$(x,y)\in D=\{(r,\theta)\;|\; -\pi/2 \le \theta \le \pi/2, 0\le r \le 2\cos\theta \}$, 

*$r_x\times r_y =(1,0,-2x)\times (0,1,-2y)=(2x,2y,1)$, which indeed is correctly oriented.


It follows that your flux equals
$$
\phi=\iint_{\Sigma}\vec{F}\cdot d\vec{S}=\iint_{D}(0,x,0)\cdot(2x,2y,1) dA=\iint_{D}2xy\; dA=\int_{-\pi/2}^{\pi/2}\int_0^{2\cos\theta}2r^3\cos\theta\sin\theta\;drd\theta,
$$
which yields $\phi=0$.
Method 2 Use your second parametrization (in cylindrical coordinates). It will give you the exact same integral 
$$\int_{-\pi/2}^{\pi/2}\int_0^{2\cos\theta}2r^3\cos\theta\sin\theta\;drd\theta.
$$
I think it is slightly better to use cartesian coordinates for parametrization as computing $r_x \times r_y$ is always easy if $x$ and $y$ are your parameters. It may not always be the case with parameters $r$ and $\theta$.
Method 3 
Note that your surface $\Sigma$ is an open surface, but it is possible to close it with the plane $z=16-2x$. Indeed, the cylinder $x^2+y^2=2x$ intersects $z=16-x^2-y^2$ at $z=16-2x$. Let $S$ be the portion of this plane that closes $\Sigma$ underneath. A normal unitary vector of $S$ is given by $\frac{1}{\sqrt5}(2,0,-1)$. By the divergence theorem:
$$
\phi = \iint_{\Sigma\cup S}\vec{F}\cdot d\vec{S}-\iint_{ S}\vec{F}\cdot d\vec{S}=\iiint_E div(\vec{F})\; dV - \iint_{ S}(0,x,0)\cdot \frac{1}{\sqrt5}(2,0,-1) \;dS=0-0=0.
$$
Note how the orientation of $\frac{1}{\sqrt5}(2,0,-1)$ (oriented towards z<0) is compatible with the divergence theorem.
The nature of the field $(0,x,0)$ kind of hints that this method should work, as the divergence will be 0, but computing the second part of the integral (the flux through $S$) can be tricky depending on the surface. But sometimes it requires no computing at all, like here!
