Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 2 / 16 / a / Solution

The cube is $1\le x\le 2$, $0\le y,z\le 1$, so $|x|=x$ throughout it. Evaluate the six outward surface integrals directly.

On $x=2$ the normal component is $2$, giving flux $2$; on $x=1$ it is $-1$, giving flux $-1$. On $y=1$ the normal component is $2x$, giving ${\int }_1^2{\int }_0^{1}2xdzdx=3$; on $y=0$ it is zero. On $z=1$ the normal component is $e^{-1}$ and on $z=0$ it is $-1$, so those unit-area faces give $e^{-1}$ and $-1$.

Adding all six contributions yieldsOnly direct surface integral was used, as requested.