## Surface Integral

Evaluate $\int {F.dS}$ where, $F=xi+(z^2-zx)j-xyk$ and $S$ is a triangular surface with vertices (2,0,0) , (0,2,0) and (0,0,4)

My attempt :

The equation of the triangular plane : $P=2x+2y+z-4$

The normal surface vector : $\nabla {P}=2i+2j+k$

Unit normal surface vector : $\frac{2i+2j+k}{3}$

Projecting the triangle plane on xy plane,

$dS=\frac{dxdy}{\frac{1}{3}}$

$\therefore S=\int_0^2{\int_0^{2-y}(2i+2j+k).(xi+(z^2-zx)j-xyk)dxdy$

As the projection is on xy plane, z=0

On integration, S=2

The correct solution however is $\frac{10}{3}$

Please let me know where my logic is wrong.

Thank you.