Evaluate $\iint\vec F\cdot\vec n\,\mathrm dS$ where $\vec F = 4zx\,\vec\imath - y^2\,\vec\jmath + yz\,\vec k$, $S$ is the surface of the cube bounded by $x=0$ ,$x=a$, $y =0$, $y=a$, $z=0$, $z=a$.
$$\iint_S\vec F\cdot\vec n\,\mathrm dS = \iint_S \frac{F\cdot\nabla(S)}{|\nabla(S)\cdot\vec p|}\,\mathrm dA$$
where $\mathrm dA$ is the element of the shadow region of the surface element $\mathrm dS$, $\vec p$ is the unit normal vector to $\mathrm dA$
$$\iint_S\vec F\cdot\vec n\,\mathrm dS = \iint_S\frac{\vec F\cdot\vec k}{\left|\vec k\cdot\vec p\right|}\,\mathrm dA$$
Then
$$\iint_S\vec F\cdot\vec n\,\mathrm dS= \iint_S\frac{yz}{\left|\vec k\cdot\vec k\right|}\,\mathrm dA= \iint_S yz\,\mathrm dA= \iint_S ay\,\mathrm dA$$
Then I'm getting the answer $\frac{a^4}2$ and since there are $6$ surfaces so the total flux should be $3a^4$. But the answer is given $\frac{3a^4}2$. Where did I go wrong ?
The flux is not the same for all faces. For example, if $z=0$ then the versor is $n=(0,0,-1)$ and $$\int_{x=0}^a\int_{y=0}^a F(x,y,0)\cdot (0,0,-1) dxdy=0.$$ In a similar way we got zero for $x=0$ and $y=0$. For $x=a$, $$\int_{y=0}^a\int_{z=0}^a F(a,y,z)\cdot (1,0,0) dydz=4a\int_{z=0}^a zdz=2a^4.$$ For $y=a$, $$\int_{x=0}^a\int_{z=0}^a F(x,a,z)\cdot (0,1,0) dxdz=-a^4.$$ For $z=a$, $$\int_{x=0}^a\int_{y=0}^a F(x,y,a)\cdot (0,0,1) dxdy=a\int_{z=0}^a ydy=\frac{a^4}{2}.$$ Therefore the total flux is $3a^4/2$.
The evaluation of the total flux is much easier if we apply the divergence theorem: the flux of $F$ through the surface of the cube is the integral of $\text{div}(F)=4z-2y+y=4z-y$ over the whole cube: $$\iiint_{[0,a]^3}(4z-y)dxdydz=4a^2\int_0^a z dz-a^2\int_0^a y dy=2a^4-\frac{a^4}{2}=\frac{3a^4}{2}.$$