Question: Calculate the flux of the vector field $\vec{F}(x,y,z)=3\vec{i}−3\vec{j}+5\vec{k}$
through a square of side length $5$ lying in the plane $4x+2y+4z=1$,oriented away from the origin.
My Solution(its wrong):
I found a square of length $5$ on the plane: $$0 \le x \le 5/(2)^{1/2}$$ and $$0 \le y \le 2\cdot(5)^{1/2}$$ (I might have accidentally reversed $x$ and $y$ but does not matter either way)
The normal vector is $(1,1/2,1)$, which is orientated away from the origin.
I take the integral of the dot product of $(3,-3,5)$ and $(1,1/2,1)$ to get $13/2$.
So the integral because nothing depends on $x$ and $y$ is just $(5/(2)^{1/2})\cdot (2\cdot(5)^{1/2})\cdot (13/2)$ which is $(65(5)^{1/2})/(2)^{1/2}.$
This answer is incorrect and I do not understand why. Would someone be able to explain where the flaw in my logic is.