Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 16 / b

The PDF diagram specifies the prism asIts triangular horizontal cross-section has area four. Calculate the outward flux on each face using its outward normal vector.

On $x=0$, the normal is $-{\mathbf{e}}_x$ and $F_x=0$, so the flux is zero. On $y=0$, the normal is $-{\mathbf{e}}_y$, givingOn the sloping vertical face $y+2x=4$, the outward normal is proportional to $(2,1,0)$. Both $F_x$ and $F_y$ vanish there, while $F_z$ is tangent to the face, so the flux is zero.

On the bottom $z=0$, the normal is $-{\mathbf{e}}_z$ and $F_z=1$, so the flux is minus the base area, namely $-4$. On the top $z=1$, $F_z=0$, so the flux is zero. Thus all five face values and their sum are

Differentiate the corresponding components:

Use the prism bounds read from the PDF diagram and the divergence from part (ii):Set $A=4-2x$ in the inner evaluation. Since $4x-9=-2A-1$, the inner integral is $A^2-(2A+1)A=-A^2-A$. Therefore

The two independently computed answers agree:This verifies the divergence theorem. Its hypotheses hold because the field is smooth and the prism has a closed piecewise smooth boundary with consistently outward normals.