A vector field is given as $A = (yz, xz, xy)$ through surface $x+y+z=1$ where $x,y,z \ge 0$, normal is chosen to be $\hat{n} \cdot e_z > 0$. Calculate the flux of the vector field.
I tried using Gauss theorem $ \iint_S A \cdot \hat{n}dS = \iiint_D \nabla \cdot A dV $, but $\nabla \cdot A $ gave the result of $0$, so I'm unsure how to tackle this problem. Any clues are welcome!
Calculating it directly
$z = 1 - x- y\\ dS = (1,1,1)$
$\int_0^1 \int_0^{1-x} (y(1-x-y),x(1-x-y), xy)\cdot(1,1,1) \ dx\ dy$
$\int_0^1 \int_0^{1-x} x+y - xy - x^2 - y^2 \ dx\ dy\\ \int_0^1 (x-x^2)(1-x)+ (1-x)(\frac 12 (1-x)^2) - \frac 13 (1-x)^3\ dy\\ \int_0^1 x(1-x)^2+ \frac 16 (1-x)^3\ dy\\ \int_0^1 (1-x)^2- \frac 56 (1-x)^3\ dy\\ \frac 13 - \frac 5{24} = \frac 18$
As you suggested using the divergence theorem.
We can create a tetrahedron 3 triangles in the xy,yz, xz planes.
$\iint f(x,y,z) \ dA_1 + \iint f(x,y,z) \ dA_2 + \iint f(x,y,z) \ dA_3 + \iint f(x,y,z) \ dS = \iint \nabla \cdot f dV$
$\nabla \cdot f = 0$
and by the symmetry of the figure.
$\iint f(x,y,z) \ dA_1 = \iint f(x,y,z) \ dA_2 = \iint f(x,y,z) \ dA_3$
$\iint f(x,y,z) \ dS = -3 \iint f(x,y,z) \ dA_1$
We need our normals pointed outward. $(0,0,-1)$ would be the be the suitable normal
$\iint f(x,y,z) \ dS = 3\int_0^1 \int_0^{1-x} xy\ dx\ dy$