Flux of vector field in a plane

Question

Compute the flow of the vector field $F(x, y, z) = (0, 0, z)$ up through the part of the plane $x + y + z = 1$ that lies in the first octant.

My take on it:

We have that $z= 1-x-y$ and $0≤y≤1-x$ and $0\le x$ but I am not sure about the upper bounds for $x$ (Can someone explain that?).

Anyways I get $(x,y,1-x-y)$ and then I partial differentiated this and got $f_x=(1,0,-1)$ and $f_y = (0,1,-1)$. Then I took the cross product of the two vectors and it is (1,1,1). So my integrand is the dot product of $(0,0,z)$ and $(1,1,1)$. This obviously equals to $z$. And the absolute value of the cross product vector is $\sqrt{3}$. I then just randomly said $x\le 1$ and got:

$$\sqrt{3}\iint_D 1-x-y \,dx\,dy = \frac{\sqrt{3}}{3}$$

Answer 1

Please read this answer why in this situation you should not normalize the vector $(1,1,1)\,.$ So there is no $\sqrt{3}\,.$ The bounds to every coordinate are $[0,1]$ which should answer one of your questions. The integral itself should be \begin{align} &\int_0^1\int_0^{1-x}(1-x-y)\,dy\,dx=\int_0^1\Big(1-x-x(1-x)-\frac{(1-x)^2}{2}\Big)\,dx\\[3mm] &=\int_0^1\frac{(1-x)^2}{2}\,dx=\frac{1}{6}\,. \end{align}