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

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.

Parameterize the surface by $\mathbf{r}(x,z)=(x,1+z,z{)}^T$. Its two tangent vectors are $(1,0,0{)}^T$ and $(0,1,1{)}^T$, whose cross product is $(0,-1,1{)}^T$ of Euclidean norm $\sqrt{2}$. Thus the scalar area element of a surface isEquivalently, projection onto the $xz$-plane reduces the surface area by $|\mathbf{n}\cdot {\mathbf{e}}_y|=1/\sqrt{2}$.

On the surface $y=1+z$, the integrand is $x+z+z^2$. Using the area element of a surface from part (i),

On the unit sphere, the outward unit normal is $\mathbf{n}=(x,y,z{)}^T$, so the normal component of the radial vector field is $\mathbf{F}\cdot \mathbf{n}=x^2+y^2+z^2=1$. The direct surface integral isHere $dS=\sin \theta d\theta d\varphi$ is the unit-sphere area element of a surface in spherical coordinates.

On the same sphere, the normal component is $(-y,x,0)\cdot (x,y,z)=-xy+xy=0$. This vector field is tangent to the sphere everywhere. Therefore its outward flux is