The flux of the vector field (z,x,y) across the surface S is −1.
Let's go through the calculation again to find the flux of the vector field (z, x, y) across the surface S, where S is the part of the plane z=x + y above the unit square 0≤x≤1, 0≤y≤1.
We want to find:

where F=(z, x, y) and dS is the outward-pointing normal vector to the surface S.
Parameterize the Surface:

Compute Partial Derivatives:

Compute the Cross Product:

Compute the Flux:

Now, integrate over the unit square:
![\begin{aligned}& \text { Flux }=\int_0^1 \int_0^1(-2 x) d y d x \\& =\int_0^1[-2 x y]_0^1 d x \\& =\int_0^1(-2 x) d x \\& =\left[-x^2\right]_0^1 \\& =-1-(0) \\& =-1 .\end{aligned}](https://img.qammunity.org/2024/formulas/mathematics/high-school/xz6ijhlmqveddetib9swb0jmttiqjbadw5.png)
Question:
Consider the surface S which is the part of the plane z= x + y, above the unit square 0≤x≤1, 0≤y≤1. Find the flux of (z, x, y).