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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.

For the parametrized surface in i,These tangent vectors are orthogonal, and their Euclidean norms are $\sqrt{(c-a{)}^2+(d-b{)}^2}$ and $\rho$, respectively. Their cross product therefore has Euclidean norm $\rho \sqrt{(c-a{)}^2+(d-b{)}^2}$. Integration givesThis is the lateral surface area of a conical frustum, with no end disks included.

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.

For a smooth surface the friction force vanishes. The ${\mathbf{e}}_{\theta }$ component of Newton's second law then gives $r^2\dot{\theta }=h$, constant. Along the constraint $z=H(r)$, the chain rule givesThe acceleration in cylindrical coordinates isTake the dot product of Newton's second law with the tangent vector ${\mathbf{e}}_r+H^{\prime }{\mathbf{e}}_z$. The normal force disappears, while the gravity term is $-mg{H}^{\prime }$. Dividing by $m$ givesUsing $\dot{\theta }=h/r^2$ yields

Continue with $f=p\sin \varphi$ and $q=p\cos \varphi$. The tangent vector to the hoop is ${\mathbf{r}}_{\varphi }=f^{\prime }{\mathbf{e}}_1-q^{\prime }{\mathbf{e}}_3$. Smooth contact means that the hoop's normal force has zero dot product with this tangent vector. The remaining applied force is $-mg{\mathbf{e}}_3$, whose dot product with ${\mathbf{r}}_{\varphi }$ is $mg{q}^{\prime }$.

Since $\mathbf{\omega }=\omega {\mathbf{e}}_3$, the Coriolis force is perpendicular to the hoop's plane and hence to ${\mathbf{r}}_{\varphi }$. AlsoTake the dot product of the rotating-frame Newton's second law with ${\mathbf{r}}_{\varphi }$, substitute a, and divide by $m$:DefineThen $K^{\prime }/2=f^{\prime }{f}^{\prime \prime }+q^{\prime }{q}^{\prime \prime }$ and $J^{\prime }=-{\omega }^{2}f{f}^{\prime }-g{q}^{\prime }$, soExpanding the squares makes the cancellation of the mixed terms explicit:The strict inequality follows from $p>0$. Here $J$ is the rotating-frame effective potential per unit mass. Multiplication of the equation by $\dot{\varphi }$ gives the useful first integral $\frac{1}{2}K(\varphi ){\dot{\varphi }}^2+J(\varphi )=\text{constant}$.

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.

Here $m{h}^2=b$, so the Binet equation reduces toThe initial conditions becomeMultiply the differential equation by $2u^{\prime }$ and use the chain rule:The initial conditions give $C=0$. Since $e^{-u}>0$, $u^{\prime }$ cannot change sign at a finite value of $u$. Thus $u^{\prime }=e^{-u/2}$ on the physical branch. Separation of variables givesThereforeThe natural logarithm must be positive because $r>0$. For the forward trajectory, $\theta \ge 0$ and $h=\sqrt{b/m}>0$, so $\dot{\theta }=h/r^2>0$: the particle travels counterclockwise while $r$ strictly decreases.

For $a=1$, it starts at $(x,y)=(1,0)$, with tangent vector proportional to $(-e^{-1/2},1)$. Its successive crossings of the positive $x$ coordinate axis have radiiIt makes infinitely many inward turns, approaching the origin as $\theta \to \infty$. There is no finite-time arrival at the origin: $dt/d\theta =r^2/h$ behaves as $1/(4h{\log }^2\theta )$, whose improper integral diverges. If extended backwards, it comes from infinity as $\theta \downarrow 2(1-\sqrt{e})$.

On the specified curve $\mathbf{r}(t)=(t,t,t^2)$, the tangent vector is $(1,1,2t)$ andTheir dot product is $t+(\lambda -\mu -4)t^2+2t^3$. Hence

Parametrize the line by $\mathbf{r}(t)=(t,t,t)$, $0\le t\le 1$. Its tangent vector is $(1,1,1)$. On this line the three force components sum to $\mu t+(1-\lambda )t+(\lambda -1-\mu )t=0$. Therefore the line integral is $\boxed{\int \mathbf{F}\cdot d\mathbf{r}=0}$ for every parameter choice.

For positive semiaxes $a,b$, the cross product of the two parametric tangent vectors gives the outward oriented surface element of an ellipsoid of revolution:Substituting the parametrization into the vector field and taking its dot product with this element givesThe middle term has zero integral over $0\le \varphi \le 2\pi$. In the other terms, ${\int }_0^{2\pi }{\cos }^2\varphi d\varphi =\pi$, ${\int }_0^{\pi }{\sin }^3\theta {\cos }^2\theta d\theta =4/15$, and ${\int }_0^{\pi }\sin \theta {\cos }^4\theta d\theta =2/5$. The directly evaluated flux is therefore

The derivative of the parabola is $dy/dx=2+2x$, giving tangent line slope $4$ at $(1,3)$. A normal line to a plane curve has the negative reciprocal slope, since its direction is orthogonal to a tangent vector. HenceThe point $(1,3)$ satisfies the equation, and the product of the two slopes is $-1$.

On this circle the radial coordinate is constant, $r=2$, so $\dot{r}=\ddot{r}=0$. The velocity is a tangent vector to the circle in the anticlockwise direction, represented by $i{e}^{i\theta }$ when $\dot{\theta }>0$. Substituting into part (b),The inward component is the centripetal acceleration; the transverse component allows the speed to change. Anticlockwise motion does not imply uniform speed or $\ddot{\theta }=0$. The preceding figure includes the required tangent arrow.

The sloping height relation on the cone is $z=b(1-r/a)$. A parametric equation of its curved surface is thereforeIts two tangent vectors are ${\mathbf{R}}_r=(\cos \theta ,\sin \theta ,-b/a)$ and ${\mathbf{R}}_{\theta }=r(-\sin \theta ,\cos \theta ,0)$. Multiplication by $dr$ and $d\theta$ produces the two infinitesimal sides requested in the paper. Their cross product is the vector area element:Its radial component is positive, so this normal vector points outwards from the solid cone. Taking its norm gives the scalar area elementThe parametric equation degenerates at the apex, but that single point contributes no area.

For $\mathbf{r}(t)=(t^2,t{)}^T$, the tangent vector is $(2t,1{)}^T$ and $\mathbf{F}(\mathbf{r}(t))=(t^3,t^4+t{)}^T$. ThusThe unequal values for the same endpoints already rule out a conservative vector field.

With $\mathbf{r}(t)=(t,t^2{)}^T$, the tangent vector is $(1,2t{)}^T$ and $\mathbf{F}(\mathbf{r}(t))=(t^3,2t^2{)}^T$. The line integral is

At a regular constrained extremum, every tangent vector to the level set $g=A$ gives zero directional derivative of $f$. Thus $\nabla f$ is perpendicular to the tangent plane, and must be a scalar multiple of its nonzero normal vector $\nabla g$. The Lagrange multiplier equations areOne solves these four equations for $x,y,z,\lambda$, evaluates $f$ at the candidates, and classifies or compares the values as required. Points where $\nabla g=0$, and any boundary or limiting cases of the feasible domain, require separate checks because the regularity argument does not cover them.

On the first segment, parameterize ${\mathbf{r}}_1(t)=(t,t,0{)}^T$, $0\le t\le 1$. The dot products with ${\mathbf{r}}_1^{\prime }=(1,1,0{)}^T$ areTheir line integrals are one and $\sin 1$, respectively.

On the second segment take ${\mathbf{r}}_2(t)=(1,1,t{)}^T$, with tangent vector $(0,0,1{)}^T$. The corresponding integrands are $a-2+2t$ and $c$, whose integrals are $a-1$ and $c$. Adding the segments yields

For $\mathbf{F}$, all three curl components vanish for every $a$. More constructively,Hence $\mathbf{F}$ is conservative for every real $a$. Its endpoint vector-field potential difference is $a$, checking both line integrals.

For the other vector field, direct partial differentiation givesIt therefore has a vector-field potential exactly when $b=0$, with no restriction on $c$:The value of $a$ is irrelevant to $\mathbf{G}$. Its rotational part for general $b$ is $b(1,1,1{)}^T\times \mathbf{r}$. Both selected paths lie in $x=y$, and this rotational part has zero dot product with every tangent vector to that plane. That explains why these paths cannot detect its nonzero curl when $b\ne 0$.

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 $|\mathbf{n}\cdot {\mathbf{e}}_y|=1/\sqrt{2}$.

Vector identity. Work in Cartesian coordinates and use the Einstein summation convention. Contraction of the two Levi-Civita symbols givesThus the convective acceleration identity isEach term is a geometrically defined vector field: the directional derivative of a vector along itself, the gradient of a scalar field, and a cross product with the curl. Equality in a Cartesian frame is equality of these vectors, so it holds in every coordinate system. When using a varying coordinate basis, derivatives of the varying basis vectors must be included; differentiating only the component functions would describe a different operation.

Spherical metric. LetDifferentiating gives the coordinate tangent vectorsTheir pairwise dot products vanish, proving that spherical coordinates are orthogonal coordinates. Their lengths are the scale factors of orthogonal coordinates:These formulas use the usual chart $r>0,0<\theta <\pi$; the origin and polar axis are coordinate singularities.

Azimuthal field. Write the speed as $U(r)$ to distinguish it from the velocity field $\mathbf{u}=U(r)e_{\varphi }$. Using the curl in spherical coordinates,Also $\frac{1}{2}\nabla |\mathbf{u}{|}^2=U{U}^{\prime }{e}_r$. Since $e_{\varphi }\times e_r=e_{\theta }$ and $e_{\varphi }\times e_{\theta }=-e_r$,The $U{U}^{\prime }$ terms cancel in the convective acceleration identity, givingThis centripetal acceleration of an azimuthal flow points toward the polar axis. A streamline is a circle of cylindrical radius $r\sin \theta$, so its acceleration is $-U^2{e}_{\rho }/(r\sin \theta )$, with $e_{\rho }=\sin \theta e_r+\cos \theta e_{\theta }$. This independent geometric argument recovers both nonzero components and explains why $U^{\prime }$ cannot appear: the particle travels at fixed $r$.