How do you parametrize the part of the plane $x+y+z=2$ that is in the first octant ($x \ge 0, y\ge 0, z\ge0$)? Can you parametrize it in the Cartesian coordinate system like this?
$x\hat{i}+y\hat{j}+(2-x-y)\hat{k}$ 
With $0\le x\le2$
$0 \le y \le 2-x$
Since it is a part of a plane and not part of a sphere, Can you also do a parametrization in the cylindrical or spherical coordinate system?
 A: Yes, your parametrization is correct (except for the $0 \le y \le 2-x-y$ part, which should be removed).
It's possible to parametrize it using cylindrical or spherical coordinates. For cylindrical coordinates, you can for instance use $z$ and $\theta$ as the parameters, with domain $0 \le z \le 2$ and $0 \le \theta \le \pi/2$, and then you can use the plane equation to determine $r$ in terms of $z$ and $\theta$. (You could also take $r$ and $\theta$ as parameters and solve for $z$, but the form of the domain is more complicated.) Likewise you can do the same for spherical coordinates, where this time your parameters are $\theta$ and $\phi$, with $0 \le \theta \le \pi/2$ and $0 \le \phi \le \pi/2$. But generally, either of these will give more complicated expressions that will make most integrals harder.

For cylindrical coordinates, in more detail: You know that $x \ge 0$ and $y \ge 0$, so $z = 2-x-y \le 2$. To express $r$ in terms of $z$ and $\theta$, we can write $x = r \cos \theta$ and $y = r\sin\theta$, and then $2 = x+y+z = z + r(\cos\theta+\sin\theta)$, giving $$r = \frac{2-z}{\cos\theta+\sin\theta}.$$
