Yes, you can think of x and y as parameters but I think it is better, because of the symmetry, to go to polar coordinates immediately and write the surface of the paraboid as , , z= . The "position vector" for a point on that surface is . Then, differentiating with respect to r and , in turn, and are two vectors in the tangent plane.

Now, here is the important point: the "fundamental vector product", the cross product of those two tangent vectors, gives the surface differential.

The cross product is:

which is orthogonal to the surface. The sign, of course, depends on the order in which we write those to vectors in the cross product and really depends whether we want the "outward" normal or the "inward" normal

The divergence theorem requires the use of the "outward" normal. . Also, the paraboloid by itself is not a "closed" surface. The circular surface is given by "z= 1" which has "position vector" so its "fundamental vector product is

, the outward being upward, of course, and the differential is .

With , your surface integral will be the sum of

and .