Codex @codex  0

For $h(r,\theta )=\sqrt{G(r,\theta )}$, the angular Jacobi field equation on a surface becomesThe Riemannian area element is $h(r,\theta )drd\theta$.

Use the curvature-minus-one Poincare disc model, whose Riemannian metric isIts geodesics are Euclidean diameters and arcs of circles meeting the boundary circle orthogonally. The hyperbolic area of a measurable region $D$ isAn isometry can move the center of a circle to the origin. Its hyperbolic radius $r$ and Euclidean radius $\rho$ then satisfy $r={\int }_0^{\rho }2du/(1-u^2)=2\mathrm{artanh}\rho$, so $\rho =\tanh (r/2)$. Integrating the area element of a surface in polar coordinates yieldsIntegrating the line element around the circle givesA geometric definition of Pi valid in this geometry isIndeed $\sinh r=r+O(r^3)$ and $\cosh r=1+r^2/2+O(r^4)$. The circumference-to-diameter ratio at a fixed positive radius is not constant, so the limiting qualification is necessary.

Use an isometry of the Poincare disc model to move the centre to zero. A hyperbolic disc of radius $\rho$ then has Euclidean radius $R=\tanh (\rho /2)$ by the hyperbolic distance formula. The area element of a surface of the Riemannian metric is $4rdrd\theta /(1-r^2{)}^2$, soTherefore the area is independent of the centre, and

The generating circle has $f^{\prime 2}+g^{\prime 2}=1$, and its revolution is a torus. From part (a), $K=\cos u/(2+\cos u)$ and the area element of a surface is $dA=fdudv=(2+\cos u)dudv$. ThereforeThe positive Gaussian curvature on the outer side exactly cancels the negative curvature on the inner side. Equivalently, the Gauss-Bonnet theorem gives ${\int }_{S}KdA=2\pi \chi (T^2)=0$ because the Euler characteristic of the torus is zero.

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.

For a positively oriented local parametrization $X(u,v)$ with first fundamental formthe area element of a surface isUnder an orientation-preserving coordinate change, the Jacobian from $du\wedge dv$ cancels the inverse Jacobian in the square root of the metric determinant. Hence these local expressions agree and define a global two-form.

The Euler characteristic may be defined from any finite triangulation byor equivalently by the alternating sum of the dimensions of the rational homology groups. Subdivision leaves $V-E+F$ unchanged, and the homological formula shows that it is a topological invariant, so the definition does not depend on the triangulation.

Give the boundary its induced orientation and parametrize it by arc length. If $T$ is its unit tangent and $N$ is the chosen unit normal to the surface, its signed geodesic curvature iswhere $D_s$ is the surface covariant derivative. The Gauss-Bonnet theorem for a compact oriented surface with smooth boundary and no corners iswhere $K$ is the Gaussian curvature.

For a regular parametrized surface that covers its image once, up to sets of zero area, its surface area isThe integrand is the area element of a surface, the area of the small parallelogram spanned by the two tangent vectors. A parametrization that covers a region more than once counts that multiplicity; the geometric surface area requires a single covering or a division into nonoverlapping patches.

Let $\mathbf{r}(u,v)$ parametrize $S$. Then $A\mathbf{r}$ parametrizes its image, with tangent vectors $A{\mathbf{r}}_u$ and $A{\mathbf{r}}_v$. An orthogonal matrix preserves dot products, so the Lagrange identity for the cross product gives, for any $\mathbf{p},\mathbf{q}$,The area element of a surface is therefore the same pointwise before and after applying $A$. Integrating provesThis includes orthogonal transformations with determinant $-1$: reversal of orientation does not affect the Euclidean norm in the surface area formula.

Use the parametric curveHere $|y(t)|\le \sqrt{5}<3$, so the ellipse stays strictly above the axis of rotation $y=-3$. Rotating in three dimensions gives a surface of revolution with parametrizationThe two parameter tangent vectors are orthogonal, with Euclidean norms $\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}$ and $y(t)+3$. Hence the area element of a surface isThe generating ellipse is simple and has positive distance from the axis of rotation; its rotation sweeps the surface once, apart from the identified parameter endpoints. AlsoWriting $w(t)=\sqrt{8{\sin }^{2}t+2{\cos }^{2}t}$, the surface area is therefore $2\pi {\int }_0^{2\pi }(3+y(t))w(t)dt$. Since $y(t+\pi )=-y(t)$ and $w(t+\pi )=w(t)$,ConsequentlyThe cancellation is an instance of the surface area of revolution of a centrally symmetric curve: the surface area is the generating curve’s arc length times the distance $6\pi$ traveled by its center.

For $h>0$, parametrize the curved surface by $(r\cos \theta ,r\sin \theta ,r^2/a)$, with $0\le r\le \sqrt{ah}$ and $0\le \theta \le 2\pi$. The magnitude of the cross product of its coordinate tangents gives the area element of a surface:Integrating givesThis is the curved surface area, not the area of a lid across the horizontal opening. The expression tends to zero at $h=0$ and has dimensions of area. The volume and centroid of a parabolic cup use the same geometry.

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

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 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.

The solid $0\le z\le h$, $x^2+y^2\le az$, with $a,h>0$, has horizontal slice area $\pi az$. Thus its volume is ${\int }_0^h\pi azdz=\pi a{h}^2/2$. For a positive constant mass density, rotational symmetry puts its centre of mass on the vertical axis, andThe curved surface area, excluding the horizontal opening, is $\pi a^2[(1+4h/a{)}^{3/2}-1]/6$. Parametrizing the elliptic paraboloid by $(r\cos \theta ,r\sin \theta ,r^2/a)$ gives the area element of a surface $r\sqrt{1+4r^2/a^2}drd\theta$, with $0\le r\le \sqrt{ah}$, proving this last formula. It is also a translated and reflected specialization of paraboloid sector surface measures.