The theorem on pointwise convergence of a piecewise smooth Fourier series gives the average of the two one-sided values of the corresponding periodic extension. Under the usual piecewise smooth interpretation of the paper's derivative assumption, the even extension is continuous at $0$ and at the joined endpoints $\pm \pi$. ThusFor the odd extension, the values on either side of zero are $-f(0)$ and $f(0)$; those on either side of $\pi$ are $f(\pi )$ and $-f(\pi )$. ThereforeThe endpoint values are also immediate because each sine term vanishes there. If $f(0)$ or $f(\pi )$ is nonzero, the Fourier sine series does not reproduce that endpoint value. Merely assuming that the derivative is locally piecewise continuous in the open interval would be weaker than the usual piecewise smooth hypothesis; the convergence assertion uses the standard endpoint regularity implicit in that hypothesis.

The Fourier cosine series and Fourier sine series use the respective formsTheir Fourier coefficients, obtained from Fourier orthogonality, areIn particular the constant term is $a_0/2={\pi }^{-1}{\int }_0^{\pi }f(t)dt$. The Fourier cosine series is the Fourier series of the even extension $f(|x|)$, made $2\pi$-periodic. The Fourier sine series is the Fourier series of the odd extension, equal to $f(x)$ for $0<x<\pi$ and $-f(-x)$ for $-\pi <x<0$, again made $2\pi$-periodic.

Choose the scalar fieldThe allowed vector-valued Stokes identity givesEvery boundary point lies on the unit circle in the $xy$-plane, so $f=1/2$ throughout the boundary. Since the line integral of $d\mathbf{r}$ around a closed curve is the zero displacement,Only the boundary value of this scalar field mattered: the interior of the surface need not be part of a sphere. Either consistent choice of orientation gives the same zero answer.

Use the outward orientation on the unit hemisphere and counterclockwise boundary orientation on the unit circle. With $\mathbf{r}(t)=(\cos t,\sin t,0)$ and $f=x=\cos t$, the vector line integral isFor the surface integral, $\nabla f=\mathbf{i}$ and the outward unit normal vector is $\mathbf{n}=(x,y,z)$, since the radius is one. The cross product isThe integral of $y$ is zero by reflection symmetry. Using the same spherical polar coordinates as in ii, with $a=1$, the other component isHenceverifying the vector-valued Stokes identity directly, component by component. The wedge symbol in the paper denotes the cross product here.

Choose the outward unit normal vector on the upper hemisphere. The curl of $\mathbf{F}=(4y,x,2z)$ isIn spherical polar coordinates, putHere $n_z=\cos \phi$ and $dS=a^2\sin \phi d\phi d\theta$. The surface integral is thereforeThe compatible boundary orientation follows increasing $t$ in $\mathbf{r}(t)=(a\cos t,a\sin t,0)$, $0\le t\le 2\pi$. On this circle,Thus the line integral iswhich agrees with the surface integral as required by the Stokes theorem.

Let $\Sigma$ be a compact, piecewise smooth oriented surface with piecewise smooth boundary, and let $\mathbf{F}$ be continuously differentiable on an open neighbourhood of $\Sigma$. The Stokes theorem statesThe boundary orientation must agree with the chosen unit normal vector by the right-hand rule. For an upward orientation on a horizontal disk, its boundary is traversed counterclockwise when viewed from above. Reversing the orientation reverses both the line integral and the surface integral.

For each of the three constant Cartesian unit vectors ${\mathbf{e}}_j$, apply the divergence theorem to the vector field $\varphi {\mathbf{e}}_j$. Its divergence is ${\partial }_j\varphi$, soThese are precisely the three components ofThus the volume integral of the gradient is the boundary surface integral weighted by the vector surface element $d\mathbf{S}=\mathbf{n}dS$. The regularity needed here is that $\varphi$ be continuously differentiable on a neighbourhood of the closure of $R$.

One sufficient version of the divergence theorem is the following. Let $R$ be a bounded region in ${\mathbb{R}}^3$ with a closed, piecewise smooth surface as its boundary, and let $\mathbf{F}$ be a continuously differentiable function with values in ${\mathbb{R}}^3$ on an open neighbourhood of the closure of $R$. With $\mathbf{n}$ the outward unit normal vector,The boundary surface integral is the total outward flux integral. If the boundary has several components, each has the outward orientation relative to the material of $R$; inner components therefore point into the holes. Edges and isolated vertices may be omitted when assigning the unit normal vector, since they have zero surface area.

Write the centroid as $\overline{\mathbf{r}}=\overline{x}\mathbf{i}+\overline{y}\mathbf{j}$. The region has reflection symmetry in the $y$-axis: replacing $\theta$ by $\pi -\theta$ changes $x$ to $-x$ and preserves $y$ and $dA$. Thus symmetry in integration givesFor the other first moment of area, use $y=r\sin \theta$ in polar coordinates:Here the last integral has the value allowed in the paper. Dividing the first moment of area by the area found in i givesThe positive vertical coordinate agrees with the fact that the entire region lies above the $x$-axis.

Use polar coordinates with $0\le r\le {\sin }^2\theta$ and $0\le \theta \le \pi$. The Jacobian determinant gives $dA=rdrd\theta$, so the area isThe trigonometric identityreduces the integral to its constant term: both cosine terms have zero integral over this interval. Therefore

Assume $h\ne 0$. The velocity in polar coordinates is $\dot{\mathbf{r}}=\dot{r}{\mathbf{e}}_r+r\dot{\theta }{\mathbf{e}}_{\theta }$. HenceTaking the dot product of the conserved Laplace-Runge-Lenz vector with $\mathbf{r}$ givesConsequently the polar equation of a conic isOnly angles for which the denominator is positive describe the physical trajectory. Its eccentricity is $A/(mk)$, giving an ellipse, parabola, or hyperbola when this ratio is respectively less than, equal to, or greater than $1$.

If $A=0$, the angle $\gamma$ is undefined, but the same dot product identity gives the circular orbit $\boxed{r=m{h}^2/k}$. If $h=0$, the trajectory is radial and the displayed division degenerates; for $k>0$, $\mathbf{A}=-mk{\mathbf{e}}_r$ and $\gamma =\pi$ on a fixed ray, so the identity does not determine $r$ as a function of a varying angle.

With $\alpha =0$, a gives constant $h$, and the polar coordinates basis satisfies ${\dot{\mathbf{e}}}_r=\dot{\theta }{\mathbf{e}}_{\theta }$, ${\mathbf{e}}_r\times {\mathbf{e}}_z=-{\mathbf{e}}_{\theta }$. Also $\ddot{\mathbf{r}}=(F_c/m){\mathbf{e}}_r$. Differentiating the proposed Laplace-Runge-Lenz vector therefore givesFor this to vanish for arbitrary $h$, including nonzero $h$, the central force must satisfyAttraction requires $k>0$. The condition “for any value of $h$” matters: with $h=0$ alone, the Laplace-Runge-Lenz vector is constant on each radial ray for any central force, so that special case would not determine the force law.

Let $B=A^{T}A$. It is a symmetric matrix, and ${\mathbf{v}}^{T}B\mathbf{v}=\Vert A\mathbf{v}{\Vert }^2\ge 0$. By the spectral theorem for real symmetric matrices, choose an orthonormal basis ${\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_3$ of eigenvectors of $B$, with nonnegative eigenvalues ${\lambda }_1,{\lambda }_2,{\lambda }_3$.

For each $i\ne j$, take the unit parallelogram parametrized by $s{\mathbf{v}}_i+t{\mathbf{v}}_j$, $0\le s,t\le 1$. Its surface area is $1$. Moreover,The image parallelogram consequently has area $\sqrt{{\lambda }_i{\lambda }_j}$. By the assumption this equals $1$ for all three pairs, givingNo eigenvalue can be zero. Dividing the equations and using positivity yields ${\lambda }_1={\lambda }_2={\lambda }_3=1$. HenceThus testing all surfaces is stronger than necessary: the three unit parallelograms in these principal directions already determine the result. This is the three-dimensional characterization of linear surface-area isometries.

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.

The endpoint radii are $5$ and $13$, but substituting those two numbers in the finite conical frustum formula is invalid. A rotation about the $z$-axis preserves the horizontal angle between the endpoint position vectors; here $(3,4)$ and $(12,5)$ are not parallel. The radius along the generating segment is not linear in height.

Parametrize the actual segment byIts rotated circle at height $1+6t$ has radiusThus the rotated image is the surface of revolution $\mathbf{R}(t,\theta )=(\rho (t)\cos \theta ,\rho (t)\sin \theta ,1+6t)$. Applying the conical frustum formula to infinitesimal pieces of this radial profile, or directly using the cross product as in ii, gives the surface area of a surface of revolutionTo evaluate it explicitly, put $u=82t+31$. Then $9676t^2+7316t+1861=(59u^2+19602)/41$. DefineThe chain rule gives $G^{\prime }(u)=2\sqrt{59u^2+19602}$, so the answer isThe positive logarithm arguments make this a real-valued expression. Rotating the segment's endpoints independently into one vertical plane before joining them would instead create a different surface, with surface area $180\pi \simeq 565.4867$.

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.

Interpolate the radius and height byA parametrized surface for the conical frustum isThe strict positivity of $a,c$ gives $\rho (t)>0$. Since $d>b$, the height identifies $t$ uniquely; the angle then identifies a point of each circle uniquely modulo $2\pi$. Hence this covers the lateral surface of revolution once. If $a=c$, it is the lateral surface of a circular cylinder.

Represent the quadratic form by the symmetric matrixThe spectral theorem for real symmetric matrices gives an orthogonal matrix $Q$ and real eigenvalues ${\lambda }_1,{\lambda }_2$ such that $Q^{T}MQ=\mathrm{diag}({\lambda }_1,{\lambda }_2)$. In the orthogonal change of coordinates $\mathbf{x}=Q\mathbf{u}$, the equation becomes ${\lambda }_1{u}_1^2+{\lambda }_2{u}_2^2=1$.

If both eigenvalues are positive, this is an ellipse with semiaxes ${\lambda }_1^{-1/2}$ and ${\lambda }_2^{-1/2}$. Conversely, a zero eigenvalue makes the set either empty or unbounded along that coordinate, while two negative eigenvalues give the empty set. Opposite signs give an unbounded hyperbola. A negative eigenvalue together with a zero one also gives the empty set. None of those cases is an ellipse. Therefore the exact condition is that $M$ be a symmetric positive-definite matrix.

The sum and product of its eigenvalues are its matrix trace and determinant:Two real numbers are both positive precisely when their sum and product are positive. ThusAn orthogonal transformation preserves area. Using i gives the quadratic-form ellipse area

The area of an ellipse with positive semiaxes $a,b$ is

The altitude of a triangle through $A$ has equation $(\mathbf{a}-\mathbf{d})\cdot (\mathbf{b}-\mathbf{c})=0$, and the altitude of a triangle through $B$ has equation $(\mathbf{b}-\mathbf{d})\cdot (\mathbf{a}-\mathbf{c})=0$. Their normal vectors $\mathbf{b}-\mathbf{c}$ and $\mathbf{a}-\mathbf{c}$ are linearly independent because the triangle is nondegenerate. Consequently these two lines meet in exactly one point $D$.

Substitute its position vector $\mathbf{d}$ in the supplied identity. Its first term vanishes by the first altitude of a triangle equation, and its second term vanishes by the second altitude of a triangle equation. HenceThis is exactly the condition that $D$ lies on the altitude of a triangle through $C$. The three lines therefore meet in a unique point, the orthocenter. This argument includes right triangles and triangles whose orthocenter lies outside them.