I am trying to find the flux of $F = (x, y, z)^T$ through the surface $A = \{(x, y, z) \in \mathbb{R}^3\mid 0 \leq z \leq 1 - x^2 - y^2\}$. To my knowledge the surface can be broken up to parts $A_1 = \{(x, y, z) \in \mathbb{R}^3\mid z = 0, x^2 + y^2 \leq 1\}$ and $ A_2 = \{(x, y ,z) \in \mathbb{R}^3 \mid x^2 + y^2 \leq 1 - z\}$. Right now I am struggling with 1.) finding normal vector to $A_2$ 2.) calculating the corresponding flux. The hint I have been given is that $A_2$ should be parametrized as the plot of a function. The issue is that I don't know how to apply the hint. Do we want to define $f(x, y) = 1 - x^2 - y^2$ and then, something? How does this help us to find the normal vector for $\int\int_{A_2}F \cdot n dS$?
Moreover, to my knowledge the normal vector for $A_1$ is $-k$, so that $F \cdot (-k) = -z$. Then won't the flox through $A_1$ be zero, as $\int\int_{A_a}F \cdot n dS = \int\int_{A_1}-zdS$?
In cartesian coordinates, the surface can be parametrized as $z = f(x,y) = 1 - x^2 - y^2, 0 \leq z \leq 1$. The outward normal vector to the surface is then $(-\frac{\partial z}{\partial x}, -\frac{\partial z}{\partial x}, 1) = (2x, 2y, 1)$. Note the projection of the surface in $XY$ plane is $x^2+y^2 \leq 1$
So the surface integral is $ \ \displaystyle \iint_{x^2+y^2 \leq 1} \vec F \cdot (2x, 2y, 1) \ dx \ dy$
Another way to calculate flux will be to close the surface with unit disk at $z = 0$ and apply divergence theorem. You have to then subtract the flux through the disk to find flux through the paraboloid surface (by the way, in this case, the flux through the disk at $z = 0$ is zero).