
Surface Integral
Evaluate $\displaystyle \int {F.dS}$ where, $\displaystyle F=xi+(z^2zx)jxyk$ and $\displaystyle S$ is a triangular surface with vertices (2,0,0) , (0,2,0) and (0,0,4)
My attempt :
The equation of the triangular plane : $\displaystyle P=2x+2y+z4$
The normal surface vector : $\displaystyle \nabla {P}=2i+2j+k$
Unit normal surface vector : $\displaystyle \frac{2i+2j+k}{3}$
Projecting the triangle plane on xy plane,
$\displaystyle dS=\frac{dxdy}{\frac{1}{3}}$
$\displaystyle \therefore S=\int_0^2{\int_0^{2y}(2i+2j+k).(xi+(z^2zx)jxyk)dxdy$
As the projection is on xy plane, z=0
On integration, S=2
The correct solution however is $\displaystyle \frac{10}{3}$
Please let me know where my logic is wrong.
Thank you.