Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 2 / 17 / b

Use the two pieces of the surface integral directly; this avoids introducing an unnecessary second derivative of $f$. On the unit sphere, $\mathbf{n}=\mathbf{r}$ and $\mathbf{F}\cdot \mathbf{n}=f^{\prime }(z)$. A spherical band between heights $z$ and $z+dz$ has area $2\pi dz$, so the curved flux isby the fundamental theorem of calculus. On the horizontal top disk, $\mathbf{F}\cdot \mathbf{n}=z_0{f}^{\prime }(z_0)$ and the area is $\pi (1-z_0^2)$. ThusAs usual for these surface integrals, the first derivative is assumed sufficiently integrable for the displayed fundamental theorem of calculus evaluation; a continuous first derivative suffices. Mere pointwise differentiability without such an integrability assumption need not make an ordinary flux integral exist.

Chooseor add any constant. Then $f^{\prime }(z)=1$, so the vector field is exactly $\mathbf{r}$ and necessarily has the same flux as in part (a). The formula from part (b)(i) confirms this: