Flux through Square on Plane

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.

• I formatted the question a lot, please check if I didn´t do anything unintended – Vinyl_cape_jawa Feb 24 at 17:48
• Everything looks great, thanks! – jts 307 Feb 24 at 17:49
• A parallel projection does not preserve angles, the region in the plane $4 x + 2 y + 4 z = 1$ corresponding to $0 < x < a \land 0 < y < b$ is not a square. The point is that you do not need to find a parametrization. Write the flux as $\int_D \mathbf F \cdot \hat {\mathbf n} \,dS$, where $\hat {\mathbf n}$ is the unit normal with the specified orientation. – Maxim Feb 24 at 20:40
• Thank you, so I basically just take the normal of the plane <4,2,4> and reduce it to <2/3,1/3,2/3> and go from there. – jts 307 Feb 24 at 20:56