Evaluate the flux of the vector field $\vec F = -9\hat j- 3 \hat k$ on the surface $z=y$ bounded by the sphere $x^2+y^2+z^2=16$ 
Evaluate the flux of the vector field $\vec F = -9\hat j- 3 \hat k$ on the surface $z=y$ bounded by the sphere $x^2+y^2+z^2=16$

My attempt:
$$\iint_S \vec F \cdot \vec n dS = \iint_S (0,-9,-3) \cdot (0,1,-1) dS = -6\iint_S  dS = -6A$$
Where $A$ is the area of the surface.
$A$ equal to the area of a circle with radius $4$, so $A= \pi \cdot 4^2 = 16 \pi$
Therefore the flux is:
$$\iint_S \vec F \cdot \vec n dS = -6A = -96\pi$$
But the correct answer is $-48 \sqrt{2}\pi$.
Where is my mistake?
 A: $\vec{n}$ should have a unitary norm, so you need to divide your answer by $\sqrt{2}$ (which is equivalent to multiplying by $\frac{\sqrt{2}}{2}$).
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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The 'intersection' is given by
$\ds{16 = x^{2} + y^{2} + y^{2} = x^{2} + 2y^{2}\ \imp
1 = {x^{2} \over \color{#f00}{4}^{2}} + {y^{2} \over \pars{\color{#f00}{2\root{2}}}^{2}}}$. The area you are looking for is given by $\ds{\pi \times \color{#f00}{4} \times \color{#f00}{2\root{2}} = 8\pi\root{2}}$ which yields
$$
\pars{8\pi\root{2}}\pars{-6} = \color{#f00}{-48\pi\root{2}}
$$
