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Length functional. On the unit sphere, write the position vector in spherical coordinates asIts two coordinate tangent vectors satisfy $r_{\theta }\cdot r_{\varphi }=0$, $|r_{\theta }|=1$ and $|r_{\varphi }|=\sin \theta$. Consequently the induced Riemannian metric is $d{s}^2=d{\theta }^2+{\sin }^2\theta d{\varphi }^2$. On a segment for which $\theta$ increases and $\varphi =\varphi (\theta )$,Segments where $\theta$ is not monotone require another parameter or a subdivision. Endpoint turning points can be handled by a limit.

First integral. With $F=\sqrt{1+{\sin }^2\theta ({\varphi }^{\prime }{)}^2}$, there is no explicit $\varphi$ dependence. The Euler-Lagrange equation therefore says thatis constant. For $c\ne 0$, put $k=1/|c|$ and solve:The radicand must be nonnegative along the chosen segment. The printed expression with both limits fixed at ${\theta }_1,{\theta }_2$ describes a longitude difference between endpoints; a formula for $\varphi (\theta )$ needs the variable upper limit and the additive constant shown here. The case $c=0$ gives a meridian with constant longitude. The equator is also a stationary route, although it cannot be parameterized using changing $\theta$.

Integration and geometric meaning. Put $t=\cot \theta$, so $d\theta =-dt/(1+t^2)$ andFor a nonconstant segment $k>1$, and writing $K=\sqrt{k^2-1}$, the integral becomesRearranging, and absorbing phase and sign into two constants, givesMultiplying by $\sin \theta$ shows that the corresponding Cartesian coordinates satisfy $z=Ax+By$. Thus every such stationary route lies in a plane through the sphere's center: it is a great circle, the geodesic of the sphere. The meridians and equator are the limiting or separately parameterized cases of the same description.

Rome to Singapore. Use exactly the approximate coordinates supplied in the question: $(\theta ,\varphi )=(\pi /4,\pi /12)$ at departure and $(\pi /2,7\pi /12)$ at arrival. Inserting them into the preceding relation determinesIndeed the right side is one at Rome and zero at Singapore. This corresponds to $K=1$, $k=\sqrt{2}$. In latitude $\ell =\pi /2-\theta$, the route is $\ell =\arctan [\cos (\varphi -\pi /12)]$: it starts at $45^{\circ }$ north with an initially eastward heading and then curves southward to the equator. The first-integral expression in $\theta$ has an integrable endpoint singularity at departure, where $\theta$ has a turning point; the longitude parameter is regular there.

For an explicit regular arc-length parametrization, let $a=r(\pi /4,\pi /12)$ and $b=r(\pi /2,7\pi /12)$. Their dot product is $\sin (\pi /4)\cos (\pi /2)=0$. HenceBoth $|r(s)|$ and $|r^{\prime }(s)|$ equal one. The shorter great circle arc is the global shortest path: for any unit-speed curve from $a$, its central angle $\alpha =\arccos (a\cdot r)$ satisfies $|d\alpha /ds|\le 1$ by the Cauchy-Schwarz inequality, since the tangential projection of $a$ has length $\sin \alpha$. Integrating gives length at least the endpoint angle $\pi /2$, which this route attains. For a sphere of radius $R$, the corresponding shortest distance is $\pi R/2$.

Coordinate surfaces and sketch. Adding and subtracting the defining relations gives $s=u^2+v^2$ and $x=u^2-v^2$. Since $y^2=s^2-x^2$, we have $y^2=4u^2{v}^2$. For $u^2=c>0$, eliminate $v^2$ to obtainThis is a parabolic cylinder parallel to the $z$ axis, opening towards negative $x$, with cross-sectional vertex $(c,0)$ and focus at the origin. Similarly, $v^2=d>0$ givesa parabolic cylinder opening towards positive $x$, with vertex $(-d,0)$ and the same focus. The degenerate $u^2=0$ surface is the half-plane $y=0,x\le 0$; $v^2=0$ gives $y=0,x\ge 0$. Constant $w$ gives the horizontal planes $z=w$.

Thus the requested cross-sections consist of $x=1-y^2/4$, $x=2-y^2/8$, $x=y^2/4-1$, $x=y^2/8-2$ and the two horizontal half-axes. The parabolas of the two families meet orthogonally.

Signed coordinates. Take $u\ge 0$ and choose the sign of $v$ to agree with $y$ away from $y=0$. Then the Cartesian coordinates areEvery coordinate triple gives exactly one point. For $u>0$, $v$ and $y$ have the same sign. A fixed signed $v\ne 0$ selects the upper or lower half of its parabolic cylinder; the entire cylinder corresponds to fixed $v^2$. On the cut $u=0$, the two values $v=\pm \sqrt{-x}$ represent the same point of the negative $x$ half-axis. This boundary identification is unavoidable: there $y=0$ even when $v\ne 0$, so the sign prescription is understood by approaching from either side of the cut. The interior $u>0$ gives a unique coordinate chart off that half-axis.

Orthogonality and metric. Differentiate the position vector:The pairwise dot products vanish. Their lengths give the scale factors of orthogonal coordinates and hence the Riemannian metric:These are parabolic cylindrical coordinates. The differential calculation is regular for $u^2+v^2>0$; the $z$ axis is the coordinate singularity where the two planar tangent vectors vanish.

Laplacian. Put $h=2\sqrt{u^2+v^2}$. The physical components of the gradient of a $w$-independent scalar field are ${\varphi }_u/h,{\varphi }_v/h,0$. Applying the divergence formula cancels the two equal scale factors of orthogonal coordinates inside the derivatives:Therefore, away from the coordinate singularity,

For $a>0$, the positive constant-$u$ surfaces in elliptic cylindrical coordinates satisfyThey are confocal elliptic cylinders with focal lines $x=\pm a,y=0$. For a generic fixed $v$, eliminating $u$ givesBecause $u\ge 0$, a fixed $v$ selects the half-branch with the signs of $x$ and $y$ fixed by $\cos v$ and $\sin v$. These are hyperbolic cylinders. The degenerate $u=0$ surface is $y=0$, $|x|\le a$, extended in $z$. At $v=0,\pi$ the cross-sections are the two exterior horizontal rays; at $v=\pi /2,3\pi /2$ they are vertical half-axes.

Differentiate the position vector to get the coordinate tangent vectorsTheir pairwise dot products vanish. Their squared lengths are $a^2({\sinh }^{2}u+{\sin }^{2}v)$, $a^2({\sinh }^{2}u+{\sin }^{2}v)$ and one. This proves orthogonality and gives the scale factors of orthogonal coordinates:The Riemannian metric is $d{s}^2=h^2(d{u}^2+d{v}^2)+d{z}^2$. Away from the focal lines, divide the coordinate tangent vectors by their lengths to obtain the local orthonormal unit vectors $e_u,e_v,e_z$. For physical vector field components $A_u,A_v,A_z$, the formulas for vector calculus in elliptic cylindrical coordinates areThe gradient of a scalar function has components $({\varphi }_u/h,{\varphi }_v/h,{\varphi }_z)$. Apply the divergence formula to this gradient; Laplace's equation becomesThe formulas apply where $h>0$; the focal lines require a limit or another coordinate chart.

The coordinate tangent vectors are the partial derivatives of the position vector:Their pairwise dot products vanish. Their lengths, the scale factors of orthogonal coordinates, are $1,r,1+r\sin \alpha$. Since $R<1$, the last is positive throughout the body. Thus these circular torus coordinates are orthogonal wherever $r>0$; at $r=0$ the cross-section angle degenerates, as in ordinary polar coordinates.

The ordered Jacobian determinant isIts sign is positive for $r>0$, so the volume element is $dV=r(1+r\sin \alpha )drd\alpha d\beta$. The coordinate degeneracy on the central circle has zero volume and does not obstruct volume integration.

The three coordinate basis vectors are the partial derivatives of the position vector with respect to the three coordinates. Direct differentiation givesThe multivariable chain rule then gives $d\mathbf{x}={\mathbf{h}}_{\xi }d\xi +{\mathbf{h}}_{\theta }d\theta +{\mathbf{h}}_{\varphi }d\varphi$. These are coordinate tangent vectors, not necessarily unit vectors.

A tangent vector to $M$ at $p$ is a velocity ${\gamma }^{\prime }(0)$ of a smooth curve $\gamma :(-\epsilon ,\epsilon )\to M$ with $\gamma (0)=p$. The tangent space is the set of all such velocities:It is a $k$-dimensional linear subspace of ${\mathbb{R}}^n$; for a local parametrization $\varphi$, it is the image of the injective derivative $D\varphi$ at the corresponding parameter point.

An embedded $k$-dimensional submanifold $M\subseteq {\mathbb{R}}^n$ is locally a coordinate $k$-plane: for every $p\in M$, there are open neighbourhoods $U\ni p$, $V\subseteq {\mathbb{R}}^n$ and a diffeomorphism $\Phi :U\to V$ withFor the present $C^1$ setting, $\Phi$ and its inverse are continuously differentiable; in the smooth convention both are smooth.

A tangent vector to $M$ at $p$ is a velocity ${\gamma }^{\prime }(0)$ of a continuously differentiable curve $\gamma :(-ϵ,ϵ)\to M$ with $\gamma (0)=p$. The tangent space is the collection of all these vectors:In the local coordinates above, it is $D\Phi (p{)}^{-1}({\mathbb{R}}^k\times \{0{\}}^{n-k})$, so it is a $k$-dimensional linear subspace of ${\mathbb{R}}^n$.

The regular level set theorem says that for a $C^1$ map $g:U\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^r$, if $Dg$ has rank $r$ on $g^{-1}(c)$, then $g^{-1}(c)$ is an $(n-r)$-dimensional submanifold andHere $r=1$ and the given rank condition makes the zero set a two-dimensional submanifold. It is nonempty because $g(0)=f(0)=0$. Part (c) givesConsequentlyThere is also a direct curve proof: $g(p+t(1,1,1))=g(p)=0$, so the entire line through $p$ in this direction lies in $M$. Its velocity at zero is exactly the required tangent vector.

A sufficient regular-level condition is $\nabla f(x)\ne 0$ at every point with $f(x)=0$. The implicit function theorem then makes the zero level a two-dimensional continuously differentiable submanifold. Its tangent space isFor a minimizing pair $(a,b)$ and any tangent vector $v\in T_a{S}_f$, take a differentiable surface curve $x(s)$ with $x(0)=a$, $x^{\prime }(0)=v$. The function $\Vert x(s)-b{\Vert }^2$ has a minimum at zero, so its derivative $2(a-b)\cdot v$ vanishes. Varying the other point similarly proves orthogonality to $T_b{S}_g$.

For the particular surfaces, minimize the distance from the origin to the second surface first. Put $u=x^2$, $v=y^2$, so $u,v\ge 0$ and $z=(u-v-6)/2$. ThenLet $s=u-v$. Since $u+v\ge |s|$, this is at least $|s|+(s-6{)}^2/4$, with equality when one of $u,v$ is zero. For $s\ge 0$, this quadratic has its minimum at $s=4$, with value 5. For $s\le 0$, it decreases up to zero and has minimum9 on that half-line. The global minimum is therefore5, attained atAll points of this surface lie outside the unit sphere. For any such $b$, the distance to the sphere is $\Vert b\Vert -1$, attained at $a=b/\Vert b\Vert$ by the triangle inequality. ConsequentlyThis is a global minimization, not merely a stationary-point calculation.

A tangent vector to $M$ at $p$ is a velocity ${\gamma }^{\prime }(0)$ of a smooth curve $\gamma :(-\epsilon ,\epsilon )\to M$ with $\gamma (0)=p$. The tangent space is the set of all such velocities:It is a $k$-dimensional linear subspace of ${\mathbb{R}}^n$; for a local parametrization $\varphi$, it is the image of the injective derivative $D\varphi$ at the corresponding parameter point.

A two-dimensional embedded submanifold is a subset which near every point can be straightened by a smooth local coordinate change in the ambient space to an open piece of the coordinate plane ${\mathbb{R}}^2\times \{0\}$. Equivalently locally it has a regular smooth parametrization of rank two, with the induced local topology.

A tangent vector at $x$ is the derivative ${\gamma }^{\prime }(0)$ of a smooth curve in the submanifold with $\gamma (0)=x$. These derivatives form the two-dimensional tangent space $T_{x}M$. A normal vector is a vector orthogonal to every tangent vector. Its space is $(T_{x}M{)}^{\perp }$, which has $\boxed{\dim (T_{x}M{)}^{\perp }=1}$ in the three-dimensional ambient Euclidean space.

Ordinary component differences at neighboring points depend on how the local bases vary. The covariant derivative corrects for that variation:Each upper tensor index adds a connection term and each lower index subtracts one. A scalar has ${\nabla }_{a}f={\partial }_{a}f$, while the derivative of a tensor transforms as a tensor with one additional covariant index. The Christoffel symbols themselves are not tensor components; their inhomogeneous transformation cancels that of the ordinary derivative.

For the Levi-Civita connection,These express metric compatibility and zero torsion. Along $x^a(\lambda )$, parallel transport is the initial value problemIt preserves inner products for the Levi-Civita connection. Parallel transport of the tangent itself gives the affinely parametrized geodesic equation, ${\ddot{x}}^a+{\Gamma }_{bc}^a{\dot{x}}^b{\dot{x}}^c=0$.

Transport generally depends on the path. On a round sphere, transport a tangent vector around the triangle bounded by two meridians $90$ degrees apart and the intervening equatorial arc. All three sides are geodesics; the enclosed spherical area is $\pi a^2/2$. The returning vector is rotated by $90$ degrees, with sign set by traversal orientation. Locally this holonomy is controlled by curvature, even though the connection can be made zero at a single point. A vector can therefore have changing coordinate components while remaining parallel, and constant components do not generally mean parallel transport.

This construction compares directions at separate points through a chosen connection, rather than assuming that a curved manifold supplies a path-independent comparison.^[ref]

Use a temperature-dependent strain energy density $W(F,\vartheta )$ with different minimizing deformation gradients in the two crystal phases. Above the transition temperature its wells describe austenite; below it the wells are unions $SO(3)U_i$ of martensitic variants. Left rotational orbits reflect material frame indifference, while the crystal's discrete material symmetries relate the variants. Minimizing the integral of this elastic energy, together with loading and boundary constraints, models the change of lattice shape. Compatibility restricts which low-energy gradients can meet continuously; fine compatible mixtures can accommodate macroscopic boundary shapes.

For the jump condition, write the two affine pieces as $Ax+c_{+}$ and $Bx+c_{-}$. Continuity on the interface meansTaking differences along the interface gives $(A-B)t=0$ for every tangent vector $t$ with $t\cdot N=0$. Decompose any vector $v$ as $t+(v\cdot N)N$. ThenWith $a=(A-B)N$ this provesConversely this rank-one connection permits continuity by choosing $c_{+}-c_{-}=-ka$. A coherent twin interface is such a compatible planar boundary between different martensitic variants, possibly after a rigid rotation of one variant. The rank-one condition is therefore the geometric compatibility condition for a simple twin.

For the unrotated pair, direct subtraction givesso $R=I$ is one solution, with reference interface normal $e_1$.

Put $d=1+{\delta }^2$, $c=(1-{\delta }^2)/d$ and $s=2\delta /d$. Since $c^2+s^2=1$, the stated $\overline{R}$ is a proper rotation. Multiplication givesHere $c+\delta s=1$ and $s-\delta c=\delta$. ConsequentlyThis is the second rank-one connection, now with normal $e_3$. Because $\delta >0$, the two rotations are distinct. The permitted assertion that there are at most two proves completeness.

Finally suppose $Q{U}_1$ is rank-one connected to $\overline{R}{U}_2$, with $Q\in SO(3)$. Left multiplication of their difference by $Q^T$ preserves rank and gives $U_1-Q^T\overline{R}{U}_2$. The classification just proved forces $Q^T\overline{R}$ to be either $I$ or $\overline{R}$, hence $Q=\overline{R}$ or $Q=I$. The two requested matrices are thereforeTheir connections are explicit: the first was calculated above, and $\overline{R}{U}_1-\overline{R}{U}_2=2\delta (\overline{R}{e}_3)\otimes e_1$. If desired the second matrix is

On each half-space the deformation is affine. Continuity on the separating plane makes the two affine constants equal and gives $(A-B)t=0$ for every tangent vector $t\perp N$. Decomposing any $x$ into its tangential part and $(x\cdot N)N$ givesThis is the Hadamard jump condition. Since $B=A-a\otimes N=A(I-A^{-1}a\otimes N)$, the matrix determinant lemma givesBoth determinants are positive, hence $\boxed{A^{-1}a\cdot N<1}$.

Write $a_1={\eta }_1$, $a_2={\eta }_2$, $S=a_1^2+a_2^2$ and $\Delta =a_2^2-a_1^2\ne 0$. The two proposed angles have $c=2a_1{a}_2/S$ and $s=\kappa \Delta /S$, and $c^2+s^2=1$. Therefore the displayed rotations lie in $SO(3)$. Direct multiplication givesThis is a rank-one connection for each sign. With a unit normal one may choose $N=(1,\kappa ,0)/\sqrt{2}$ and jump vector $a=\sqrt{2}\Delta (-a_2,\kappa a_1,0)/S$. The two rotations are distinct because $\Delta \ne 0$; the permitted at-most-two result therefore proves these are all solutions. Flipping both $a$ and $N$ changes no connection.

It remains to test the material plane. Because $y(0)=0$, the affine constants vanish. Let $t=(-p_2,p_1,0)$, so the plane is spanned by $t$ and $e_3$. Both deformation gradients send $e_3$ to $a_1{e}_3$. If the plane is not the twin plane, it cuts the two half-spaces; its two image halves are coplanar exactly when the in-plane vectors $Bt$ and $At$ are parallel. Put $d=(-a_2,\kappa a_1)$ and $n=(1,\kappa )$ in their two-dimensional sections. ThenIts prefactor is nonzero, so coplanarity requires $p_1=\kappa p_2$, precisely $p\parallel (1,\kappa ,0)\parallel N$. Conversely, on that twin plane $(A-B)x=0$, so the two affine maps agree and its image is a plane. Thus $\boxed{\Pi \text{ remains plane}⟺p\parallel N}$. All other allowed planes acquire a kink along their intersection with the twin. This is the planarity test across a cubic-to-tetragonal twin.

At a point of a smooth surface, a normal section of a surface can bend by different amounts depending on its tangent vector. The principal curvatures are the largest and smallest signed normal curvatures among unit tangent vectors. Equivalently they are the two eigenvalues $k_1,k_2$ of the shape operator, with an orthonormal basis $e_1,e_2$ of corresponding principal directions. Here the labels follow the selected directions, without requiring $k_1\le k_2$. ^[ref]

For a unit tangent vector making angle $\theta$ with $e_1$, Euler formula for normal curvature givesThe normal curvature is a weighted average of the principal curvatures; the extremes occur in the principal directions. If the principal curvatures coincide, every direction in the tangent plane is a principal direction: this is an umbilical point. The Gaussian curvature and mean curvature summarize the pair by

For an outward-oriented sphere of radius $R$, both principal curvatures are $-1/R$, so the normal curvature is $-1/R$ in every direction. For an outward-oriented circular cylinder, choose $e_1$ around the circumference and $e_2$ along the axis. The principal curvatures are $-1/R,0$, giving $k_n(\theta )=-{\cos }^2\theta /R$; halfway between the principal directions, $k_n(\pi /4)=-1/(2R)$.

At the origin of the upward-oriented hyperbolic paraboloid $z=(x^2-y^2)/2$, the principal directions are the coordinate axes and the principal curvatures are $1,-1$ in the chosen length unit. Thus $k_n(\theta )=\cos (2\theta )$, with zero normal curvature at $\theta =\pi /4,3\pi /4$. Zero normal curvature in these directions coexists with Gaussian curvature $-1$: the shape operator is not zero. The mean curvature is zero at this point; this does not say that the whole hyperbolic paraboloid is a minimal surface.

Each graph uses angle $\theta$ in the tangent plane, measured from the indicated first principal direction. The sphere and circular cylinder have $R=1$; the saddle graph refers only to the origin. Reversing the unit normal reflects each normal curvature graph through zero, negates both principal curvatures and the mean curvature, and preserves their product, the Gaussian curvature.

Pseudo-arclength continuation allows both the state and the parameter to change. Write $z=(u,\lambda )$ and choose a unit vector $t_k$ in the kernel of the full Jacobian matrix $DF(z_k)$. A continuation predictor-corrector step seeks the next solution on a hyperplane perpendicular to $t_k$, a distance $h>0$ along the old tangent vector. This adds one equation to $F(z)=0$ and makes the correction a square system. The constraint approximates arc length; $h$ is the displacement projected onto the old tangent vector, rather than the exact curved distance.^[ref]

A smooth map $\varphi :M\to N$ pulls a differential form on $N$ back to one on $M$ by applying its differential $d\varphi$ to each tangent vector argument. In particular, a parametrization of a curve pulls a one-form back to an ordinary integrand on an interval.

For $\alpha =xdy$ and the parabola $\varphi (t)=(t,t^2)$, $0\le t\le 1$, substitute $x=t$ and $dy=2tdt$ to obtainThe same endpoints gave $0$ and $1$ along the two square-edge paths in the introduction, so this third answer is consistent with path dependence. By contrast, ${\varphi }^{\ast }d(xy)=d(t^3)=3t^{2}dt$ gives integral $1$, the same endpoint difference as before.

The pullback of a differential form commutes with the wedge product of differential forms and the exterior derivative: ${\varphi }^{\ast }(\alpha \wedge \beta )={\varphi }^{\ast }\alpha \wedge {\varphi }^{\ast }\beta$ and ${\varphi }^{\ast }(d\alpha )=d({\varphi }^{\ast }\alpha )$. These identities make changes of coordinates and parametrizations compatible with differentiation and integration.

It is the alternating-tensor case of pullback.

The distance on a connected Riemannian manifold is the infimum of the arc lengths of piecewise smooth curves joining the endpoints:It answers “how far must I travel while staying in this geometry?” The Riemannian metric measures an instantaneous tangent vector, whereas Riemannian distance is the global result of comparing complete routes.

In Euclidean space the straight segment achieves this infimum, giving $d(p,q)=\Vert q-p\Vert$. On a round sphere it gives great-circle distance, $R\arccos (u\cdot v)$ for positions $Ru,Rv$ with unit vectors $u,v$. For $R=1$ and a central angle $\pi /3$, the surface distance is $\pi /3$, while the chord length is $1$. On a circular cylinder, routes with different winding lead to the minimum in Riemannian distance on a circular cylinder.

The word infimum matters: a shortest route need not exist. In the Euclidean plane with the origin removed, the points $(-1,0)$ and $(1,0)$ still have Riemannian distance $2$, approached by paths detouring around an arbitrarily small circle. Every allowed route has length strictly greater than $2$, since the only Euclidean straight segment passes through the missing point. The Hopf-Rinow theorem guarantees a minimizing geodesic between any two points when a connected Riemannian manifold is complete.

A Riemannian metric is a smoothly varying positive-definite inner product $g_p$ on each tangent space $T_{p}M$ of a smooth manifold. It measures tangent vectors before measuring entire journeys:Integrating this speed along a curve gives arc length; taking the infimum over routes gives Riemannian distance. A smooth manifold together with this measuring rule is a Riemannian manifold.

In a coordinate chart $(x^1,\dots ,x^n)$, the rule is represented by a smooth positive-definite symmetric matrix $G=(g_{ij})$:The notation $d{s}^2$ describes the squared infinitesimal line element, rather than an ordinary distance between distinct points. Changing coordinates changes the matrix entries; it preserves the measured norm of the same tangent vector.

For the plane with $d{s}^2=4d{x}^2+d{y}^2$, the vector $(1,0)$ has norm $2$ and $(0,1)$ has norm $1$. The coordinate change $X=2x,Y=y$ turns this into $d{s}^2=d{X}^2+d{Y}^2$: coefficients that look different can describe the same geometry. On a circular cylinder the induced metric $R^{2}d{\theta }^2+d{z}^2$ similarly becomes Euclidean when unrolled. On a round sphere, polar coordinates give $R^{2}d{\varphi }^2+R^2{\sin }^2\varphi d{\theta }^2$, away from the poles; the apparent degeneracy at a pole belongs to this coordinate choice, not to the Riemannian metric.

The point of the construction is that measurements make sense intrinsically, even without a chosen surrounding Euclidean space. It gives a common starting point for geodesics, curvature encoded by the Riemann curvature tensor, and the Laplace-Beltrami operator. ^[ref].

Take the upward-oriented graph $z=c\sqrt{1-x^2/a^2-y^2/b^2}$ over $0\le x\le L_x$, $0\le y\le L_y$, where $a,b,c,L_x,L_y>0$, $A=L_x^2/a^2$, $B=L_y^2/b^2$ and $A+B<1$. The surface flux of the vector field $F=(-y,x,0)$ is the displayed formula. Indeed the vector surface element is $(-z_x,-z_y,1)dxdy$, soThe change of variables $u=x^2/a^2$, $v=y^2/b^2$ changes $xydxdy$ to $a^2{b}^{2}dudv/4$. Integrating $(1-u-v{)}^{-1/2}$ over $[0,A]\times [0,B]$ gives $\frac{4}{3}[1-(1-A{)}^{3/2}-(1-B{)}^{3/2}+(1-A-B{)}^{3/2}]$, proving the result. When $a=b$, the value of the vector field is a tangent vector to the ellipsoid at every point, and the surface flux vanishes pointwise.