Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 2 / 16

Put $q=x^{2}yz$. The chain rule and product rule giveSumming these partial derivatives gives the Laplacian

On the face $OBA$, $z=0$ and the outward unit normal is $-{\mathbf{e}}_z$. The flux integral is thereforeThe negative sign comes from the outward orientation, which points below the tetrahedron on this face.

On $OAC$, $y=0$ and the outward unit normal is $-{\mathbf{e}}_y$. Its flux isOn $OBC$, $x=0$ and the outward unit normal is $-{\mathbf{e}}_x$. HenceOn $ABC$, write $z=1-x-y$ over $x\ge 0$, $y\ge 0$, $x+y\le 1$. The oriented surface element is $(1,1,1{)}^{T}dxdy$, pointing away from the origin. Here $x+y+z=1$, so $\mathbf{F}=(1,1,0{)}^T$. ThusThe unnormalized normal vector in the oriented surface element already incorporates the factor $\sqrt{3}$ from the surface area; no additional factor is needed.

Add the four outward fluxes:For a check, $\nabla \cdot \mathbf{F}=2(x+y+z)$. The divergence theorem gives the same result from