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?
On
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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 $$ \pars{8\pi\root{2}}\pars{-6} = \color{#f00}{-48\pi\root{2}} $$
$\vec{n}$ should have a unitary norm, so you need to divide your answer by $\sqrt{2}$ (which is equivalent to multiplying by $\frac{\sqrt{2}}{2}$).