Codex @codex  0

Let $a$ be any constant vector and take $u=\varphi a$. The product rule for divergence gives $\nabla \cdot u=a\cdot \nabla \varphi$. The divergence theorem on a bounded region with piecewise smooth boundary, oriented outward, therefore yieldsSince $a$ is arbitrary, equality of all components provesHere $\varphi$ is continuously differentiable on a neighbourhood of the closed region.

For the side of this right circular cone, the parameter tangents are $x_r=(\cos \theta ,\sin \theta ,\sqrt{3})$ and $x_{\theta }=(-r\sin \theta ,r\cos \theta ,0)$. Their cross product in the outward order isThe sign is outward because the solid right circular cone lies at smaller cylindrical radius for fixed height. Reversing the parameter order reverses the oriented surface element.

To check the integral identity, the closed boundary must include the top Euclidean disk $z=1$, radius $1/\sqrt{3}$, as well as the curved side. For $\varphi =z^2$, horizontal components cancel on integrating $\theta$. The side contribution isOn the top Euclidean disk $\varphi =1$ and $d\mathbf{S}=e_{z}dA$, so its contribution is $(\pi /3)e_z$. The total is $(\pi /6)e_z$. Independently, the cross-section of the solid at height $z$ has area $\pi z^2/3$, and $\nabla \varphi =2z{e}_z$, givingThus the two sides agree. The curved side alone is not a closed surface and does not satisfy this volume identity. The apex has zero area; alternatively one can truncate at height $\epsilon$ and let $\epsilon \downarrow 0$, with the extra boundary contribution vanishing.

Put $h(r,\theta )=\sqrt{G(r,\theta )}$. The Jacobi equation in geodesic polar coordinates and the initial conditions from part (b) areInside a geodesic polar coordinate ball, $h>0$. If $K\le 0$, then $h_{rr}\ge 0$, so $h_r\ge 1$ and therefore $h(r,\theta )\ge r$. The Riemannian area element is $hdrd\theta$, whenceThus nonpositive Gaussian curvature makes such a ball at least as large as the Euclidean disk of the same radius.

A right circular cone has an axis through its apex and the centre of its circular base. With the axis chosen as the $z$ axis, its solid interior is $0\le z\le h$, $x^2+y^2\le (z\tan \alpha {)}^2$, where $0<\alpha <\pi /2$ is the semi-angle. Its closed boundary includes both the lateral surface and the base Euclidean disk.

The real Zernike polynomials form an orthogonal polynomial basis on the unit Euclidean disk, with radial factor $R_n^m(\rho )$ and angular factor $\cos m\varphi$ or $\sin m\varphi$. Here $0\le m\le n$ and $n-m$ is even. They decompose wavefront errors into modes useful for optical correction. Normalizations differ, so coefficients should specify the chosen convention.