1. ## Flux

$F(x,y,z) = xze^y i - xze^y j + z k$
S is part of the plan $x+y+z=1$ in first octante with orientation to down

2. by divergence theorem

$F(x,y,z) = xze^y i - xze^y j + z k$

$\int\int F.n ds=\int\int\int divF dv$

$div(F)=\frac{\partial}{\partial x}(xze^y)+\frac{\partial}{\partial y}(- xze^y)+\frac{\partial}{\partial z}z$

$div(F)=ze^y-xze^y+1$

z=1-x-y

$\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y}(ze^y-xze^y+1) dzdydx$

now just evaluate the integral

3. Here's how to do it directly.

Flux = $\int \int_{S} F \cdot N \ \mathrm{d} S = \int \int_{D} F \cdot (T_{x} \times T_{y}) \ \mathrm{d}x \ \mathrm{d}y$

Let $g(x,y) = z = 1-x-y$

The surface can be parametrized by $\Phi(x,y) = \Big (x,y,g(x,y)\Big) = (x,y,1-x-y)$

$T_{x} = \frac {\partial \Phi}{\partial x} = (1,0,-1)$

$T_{y} = \frac {\partial \Phi}{\partial y} = (0,1,-1)$

EDIT : $T_{x} \times T_{y} = (1,1,1)$

The problem wants the normal vector pointing in the other direction, or $(-1,-1,-1)$

then $F \cdot (T_{x} \times T{y}) = -z = -1+x+y$

D is the region in the xy-plane bounded by the positive x and y axes and the line y = 1-x

The final final integral is then $\int_{0}^{1} \int_{0}^{1-x} (-1+x+y) \ \mathrm{d}y \ \mathrm{d}x$

4. Originally Posted by Random Variable
Here's how to do it directly.

Flux = $\int \int_{S} F \cdot N \ \mathrm{d} S = \int \int_{D} F \cdot (T_{x} \times T_{y}) \ \mathrm{d}x \ \mathrm{d}y$

D is the region in the xy-plane bounded by the positive x and y axes and the line y = 1-x

Let $g(x,y) = z = 1-x-y$
"""""????
The surface can be parametrized by $\Phi(x,y) = \Big (x,y,g(x,y)\Big) = (x,y,1-x-y)$

$T_{x} = \frac {\partial \Phi}{\partial x} = (1,0,-1)$

$T_{y} = \frac {\partial \Phi}{\partial y} = (0,1,-1)$

$T_{x} \times T_{y} = (1,1,0)$
""""""????
The problem wants the normal vector pointing in the other direction, or $(-1,-1,0)$

but then $F \cdot (T_{x} \times T{y}) = 0$? Is that right? Anybody?

D is the region in the xy-plane bounded by the positive x and y axes and the line y = 1-x

The final final integral is then $\int_{0}^{1} \int_{0}^{1-x} 0 \ \mathrm{d}y \ \mathrm{d}x$ which of course is just zero.

can you please explain that for me in details how you parametrize like that

5. [
can you please explain that for me in details how you parametrize like that
The surface can be written as z = 1 - x - y. So an obvious parametrization (at least to me) is (x, y, 1-x-y). It I had w = 1-x-y-z, then the parametrization would be (x, y, z, 1-x-y-z).

The stuff of mine you quoted has an error in it. The answer is not zero. I fixed it above.

6. I might have made a mistake somewhere because I'm not getting the same answer as your approach.

7. Originally Posted by Amer
by divergence theorem

$F(x,y,z) = xze^y i - xze^y j + z k$

$\int\int F.n ds=\int\int\int divF dv$

$div(F)=\frac{\partial}{\partial x}(xze^y)+\frac{\partial}{\partial y}(- xze^y)+\frac{\partial}{\partial z}z$

$div(F)=ze^y-xze^y+1$

z=1-x-y

$\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y}(ze^y-xze^y+1) dzdydx$

now just evaluate the integral
Ok. Thank you