Find the surface integral of
$F(x,y,z) = x^2\hat{i}+ y^2\hat{j} -z\hat{k}$ around the triangle with vertices
$(0,0,0) , (0,2,0) , (0,0,3)$
So, the equation of that triangle is actually just the straight line
$3y +2z =6$ in $yz$ plane
now, Normal vector = $\hat{i}$
Using the formula
$\int F.\hat{n} ds = \int \int x^2 dydz$
But on $yz$ plane $x =0$ so Surface Integral is zero.
So, final answer is $0$ .
Is this correct ? Can someone please verify ?
Thank you.