Vector Field $F = <(x-y),(z),(3x)>$ and $S$ is part of the plane $z = x+y$ above the rectangle $0 \leq X \leq 2$ and $0\leq Y \leq 3$ oriented upwards.
Would the bounds for the triple integral be : $x$ from $0$ to $2$, $y$ from $0$ to $3$, and $z$ from $0$ to $5$?
A double integral is required.
$<(x-y), (z), (3x)> \cdot <-fx,-fy,1> = -x+y + (-z) +3x =-x+y-x-y+3x =x$
take double integral of $x$ with bounds $x = 0$ to $2$ and $y = 0$ to $3$