Finding flux of a vector field through a hemisphere

Question

This question is given in a (publicly shared) past exam at my university: Let S be the upper hemisphere of $x^2 + y^2 + z^2 = 4$ with normal vector pointing toward the origin, and $\vec F = z \vec x / |\vec x|$ where $\vec x = <x, y, z>$. Compute $\iint\vec F \cdot d\vec S $

The answer is given as $-8\pi$; I understand that it can be obtained by parameterizing the sphere and computing $\vec t_u \times \vec t_v = <x/z, y/z, 1>$ as the normal vector where $z = \sqrt{4 - x^2 - y^2}$, then computing the integral. However, I am confused by this as I have read that the flux across a sphere can be obtained using $\vec r / | \vec r |$ as the normal, where $\vec r = <x, y, z>$. This results in a different answer, albeit not far off from $-8\pi$. Which is the proper normal to use in a flux surface integral where S is a sphere?

Answer 1

In this case, the unit normal vector $\hat{n}$ for the sphere is simply $-\vec{x}/|\vec{x}|$ by taking the gradient of the sphere equation

$$\nabla(\vec{x}\cdot\vec{x}-4=0) \implies \vec{n} = 2\vec{x}$$

then normalizing the result and choosing to point to the origin. This means we have the chain of equalities

$$\iint_S z\frac{\vec{x}}{|\vec{x}|}\cdot\vec{dS} = \iint_S -z\:dS = \iint_S(0,0,2)\cdot\vec{dS}$$

$$=\iint_D(0,0,2)\cdot(0,0,-1)dS = -2\operatorname{Area}(D) = \boxed{-8\pi}$$

where the second integral is the scalar $dS$ surface integral and $D$ is the disk in the $xy$ plane of radius $2$ (this equality is true by using Stokes' theorem

$$\iint_{S_1} \nabla\times \vec{F}\cdot \vec{dS} = \int_C\vec{F}\cdot\vec{dr} = \iint_{S_2} \nabla\times \vec{F}\cdot \vec{dS}$$

to move the integral onto another surface entirely, so long as both surfaces share the same curve boundary $C$). In this way we never computed any integration to solve this problem, only using the formula for the area of a circle.