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.

  • $\begingroup$ I formatted the question a lot, please check if I didn´t do anything unintended $\endgroup$ – Vinyl_cape_jawa Feb 24 at 17:48
  • $\begingroup$ Everything looks great, thanks! $\endgroup$ – jts 307 Feb 24 at 17:49
  • $\begingroup$ 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. $\endgroup$ – Maxim Feb 24 at 20:40
  • $\begingroup$ 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. $\endgroup$ – jts 307 Feb 24 at 20:56

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