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

The divergence of the position vector field is $\nabla \cdot \mathbf{r}=3$. At height $z$, the horizontal cross-section is a disk of area $\pi (1-z^2)$. By the divergence theorem, the outward flux integral is three times the enclosed volume:ThereforeA direct check separates the two pieces of the surface integral. On the unit sphere, the outward unit normal is $\mathbf{r}$, so $\mathbf{r}\cdot \mathbf{n}=1$ and the curved contribution is $2\pi (z_0+1)$. On the top disk, $\mathbf{n}={\mathbf{e}}_z$, giving $\pi z_0(1-z_0^2)$. Their sum is the same flux.

Differentiate the flux:On $-1<z_0<1$, $\Phi$ is strictly increasing. Its limiting endpoint values are $\Phi (-1)=0$ and $\Phi (1)=4\pi$, and $\Phi (0)=2\pi$. It has horizontal limiting tangents at the endpoints, curves upward for $z_0<0$, curves downward for $z_0>0$, and has an inflection point at zero.

The derivative graph is a downward-opening parabola, approaching zero at both endpoints and attaining its maximum $3\pi$ at $z_0=0$.

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: