Codex @codex  0

For a bounded region $R\subseteq {\mathbb{R}}^3$ with piecewise smooth boundary and a continuously differentiable vector field $\mathbf{F}$ on a neighborhood of its closure, the divergence theorem sayswhere $\mathbf{n}$ is the outward unit normal vector. The left side integrates divergence over the volume; the right side is the outward flux.

Parametrize the surface of revolution byAn oriented vector surface area element isUse the vector field $\mathbf{F}=(x/2,y/2,0)$, whose divergence is one. If this parametrization has the outward orientation on an enclosing surface, the divergence theorem givesThe second equality is integration by parts, using the endpoint values. For example this outward convention applies when the positive-$t$ branch lies above the negative-$t$ branch, $f(t)>f(-t)$ for $0<t<\pi /2$.

The stated $C^1$ regularity and endpoint conditions alone do not guarantee an embedded enclosing surface: branches can cross or coincide. Without an appropriate orientation assumption the flux formula is an algebraic signed volume. If “inside” is defined as the region between the two graph branches for each radius, its nonnegative geometric volume is insteadThis explicitly handles a reversal or change in the order of the two branches.

Write $\times$ for the paper's $\wedge$. In a right-handed Cartesian coordinate system, the cross product isIt is perpendicular to both vectors, has norm $|\mathbf{a}||\mathbf{b}|\sin \varphi$, where $\varphi$ is their angle, and has direction given by the right-hand rule when nonzero. Its norm is the area of the parallelogram spanned by the two vectors.

If $\mathbf{b}\times \mathbf{c}\ne 0$, use that parallelogram as the base of the parallelepiped. Its perpendicular height is $|\mathbf{a}\cdot \mathbf{n}|$, where $\mathbf{n}=(\mathbf{b}\times \mathbf{c})/|\mathbf{b}\times \mathbf{c}|$. Thus its volume isIf $\mathbf{b}\times \mathbf{c}=0$, both the volume and this scalar triple product vanish. If the cross product is nonzero, the scalar triple product vanishes exactly when $\mathbf{a}$ lies in the plane spanned by $\mathbf{b},\mathbf{c}$. If it is zero, $\mathbf{b},\mathbf{c}$ are linearly dependent, and adjoining $\mathbf{a}$ still spans at most a plane. Consequently the three vectors are coplanar if and only if $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=0$, including cases involving a zero vector.

Differentiate the ellipsoidal parametrization. Its outward vector surface element isFor $a,b,c>0$, its dot product with $\mathbf{r}$ simplifies to $abc\sin \theta$. Thus the requested direct surface integral isAs a check, the divergence of $\mathbf{r}$ is three, and the ellipsoid has volume $4\pi abc/3$, so the divergence theorem gives the same value.

Use a Lagrange multiplier to maximize $V=\pi r^{2}h$ subject to $2\pi r(r+h)=A$, with $r,h>0$. The equations $\nabla V=\lambda \nabla A$ areThe second equation gives $\lambda =r/2$. Substituting into the first gives $2rh=r(2r+h)$, hence $h=2r$. The constraint then gives $A=6\pi r^2$, soThis is the same aspect ratio as the minimum-area closed cylinder at fixed volume. To confirm maximality directly, eliminate $h$: $V(r)=Ar/2-\pi r^3$ on $0<r<\sqrt{A/(2\pi )}$. Its only stationary point is the value above, its second derivative there is negative, and the volume tends to zero at either end of the interval.

Use the spherical polar coordinates volume element $dV=r^2\sin \theta drd\theta d\varphi$. Since $1+\cos \theta \ge 0$ throughout $0\le \theta \le \pi$, the solid has radial range $0\le r\le 1+\cos \theta$ and full azimuthal range $0\le \varphi <2\pi$. Its volume isPut $u=1+\cos \theta$, so $du=-\sin \theta d\theta$ and the endpoints become $2$ and $0$. ConsequentlyThe polar equation describes the entire three-dimensional surface, so the spherical polar coordinates volume element, rather than a two-dimensional polar area element, is required.

The mass density becomes $\rho =\mu (2+\cos \theta )$. Integrate it over the sphere using the spherical polar coordinates and volume element above:The angular integral is $4$, since ${\int }_0^{\pi }\cos \theta \sin \theta d\theta =0$. ThereforeThe term proportional to $z/r$ contributes zero by reflection across the equatorial plane. Thus the answer is also the volume $4\pi a^3/3$ times the average mass density $2\mu$. The undefined direction at the origin affects a single point and has no effect on this volume integral.

For a bounded volume $V$ with piecewise smooth closed boundary $\partial V$, oriented by the outward unit normal vector, and a continuously differentiable vector field on a neighborhood of $\overline{V}$, the divergence theorem statesFor an oriented piecewise smooth surface $S$ with piecewise smooth boundary $C=\partial S$, and a continuously differentiable vector field on a neighborhood of $S$, Stokes theorem statesThe boundary orientation follows the right-hand rule relative to the chosen normal vector. Reversing that orientation reverses both sides of the latter identity.

The volume of a parallelepiped is the absolute scalar triple product of its edge vectors. HereThe absolute value removes the orientation of the chosen edge ordering.

With $s=\sin {\theta }_0$ and $c=\cos {\theta }_0$, conversion from spherical coordinates gives $\mathbf{a}=r_0(s,0,c)$, $\mathbf{b}=r_0(-s/2,\sqrt{3}s/2,c)$ and $\mathbf{c}=r_0(-s/2,-\sqrt{3}s/2,c)$. The volume of the parallelepiped is the absolute value of their scalar triple product:The absolute value matters when the polar angle is in the lower hemisphere. At the poles or equator the three edge vectors are linearly dependent and the volume vanishes.

For a bounded region $V$ with piecewise smooth closed boundary $S$, outward unit normal vector $\mathbf{n}$ and continuously differentiable vector field $\mathbf{F}$, the divergence theorem states ${\int }_S\mathbf{F}\cdot \mathbf{n}dS={\int }_V\nabla \cdot \mathbf{F}dV$. Here $dS$ is scalar surface area, $dV$ is volume, and $d\mathbf{S}=\mathbf{n}dS$ is the oriented area element.

For $\mathbf{F}=\mathbf{r}$, its divergence is three. The ellipsoid has semiaxes $a,a,b$, so its volume is $4\pi a^{2}b/3$, obtained by scaling the unit ball. Consequently

An axial section gives similar right triangles. At height $x$ above the base, the cone's remaining height is $h-x$, so the cylinder's radius is $r=(h-x)\tan \alpha$. Therefore its volume isAssume $h>0$ and $0<\alpha <\pi /2$ for a nondegenerate cone. The geometry and the dimensionless optimization curves are illustrated below.

Fix the total area at $A_0>0$. Apply Lagrange multipliers to $V=\pi a^2\ell$ with constraint $G=2\pi a(a+\ell )=A_0$. The two derivative equations areSince $a>0$, the second gives $\lambda =a/2$. Substitution in the first gives $\ell =2a$. The constraint then yields $a^2=A_0/(6\pi )$. ThereforeFor completeness, eliminating $\ell$ makes $V=(A_0/2)a-\pi a^3$ on $0<a<\sqrt{A_0/(2\pi )}$. Its derivative changes from positive to negative at the calculated radius and its volume tends to zero at both domain endpoints. Thus this stationary cylinder is indeed the maximum. It is the fixed-area case of optimal aspect ratios of closed cylinders.

The divergence isFor any closed piecewise smooth surface bounding a volume, use the divergence theorem with the outward orientation:Reversing the orientation still gives zero. The field is smooth everywhere, so there is no hidden singular source inside the surface.

The projection onto the $x$-$y$ plane is the disk $x^2+y^2\le 4$. In polar coordinates, the height is $4-r^2$ and the polar area element is $rdrd\theta$. The volume is thereforeThe radial factor here is required because this integral is being constructed from the geometric volume, unlike the already specified integral in (a).

Take positive dimensions measured in metres. The volume is $V=lwh$ and the external surface area constraint is $S=lw+2lh+2wh=12$. There is one base and four walls, with no lid. The Lagrange multiplier equations for $V-\lambda (S-12)$ areThe first equation implies $\lambda >0$. Multiply the first by $l$ and the second by $w$ and subtract; this gives $2\lambda h(l-w)=0$, hence $l=w=s$. The last equation gives $\lambda =s/4$. Substituting into the first yields $sh=(s/4)(s+2h)$ and therefore $s=2h$.

The constraint now reads $s^2+4sh=3s^2=12$. ThusFor a global check, apply the arithmetic-geometric mean inequality to $lw$, $2lh$ and $2wh$, whose sum is $12$. Their product is $4V^2\le 4^3=64$, so no feasible box has $V>4$. Equality holds exactly at the dimensions found above. This is the maximum-volume open rectangular box at fixed surface area.

The volume of a parallelepiped is the absolute scalar triple product of its three edge vectors. Since $\overrightarrow{OP}\times \overrightarrow{OQ}=(0,0,3)$,The factor $1/6$ appropriate to a tetrahedron must not be applied to this parallelepiped.

For positive constant mass density ${\rho }_0$, reflection symmetry makes the horizontal coordinates of the centre of mass zero. The total mass and the vertical first moment areTheir ratio givesThis is the centre of the liquid volume, rather than of the curved wall. The density cancels because it is uniform; the larger disks near the top place the centre above half-height.

Take a face on the flat base and let $x,y$ be its positive half-widths, with height $z>0$. The upper vertices meet the spherical boundary at $x^2+y^2+z^2=a^2$, and the rectangular parallelepiped has volume $V=4xyz$. A Lagrange multiplier givesMultiplication by $x,y,z$ respectively shows $x^2=y^2=z^2$. Positivity and the constraint give $x=y=z=a/\sqrt{3}$. ThereforeBoundary choices with a zero dimension give zero volume, so this is the maximum in this configuration.

The maximum-volume cuboid in a hemisphere also has a bound covering arbitrary orientations. Let $u_i$ be any box's half-edge lengths, $v_i$ its centre coordinates in its orthonormal edge basis, and $n_i$ the vertical unit-vector coordinates. Remaining above the base plane requires $\sum n_i{v}_i\ge \sum |n_i|u_i$, so $|v_j|\ge u_j$ for at least one index. Its farthest vertex obeyswhere containment supplies the first inequality. The arithmetic-geometric mean inequality applied to the three terms $u_i^2$ for $i\ne j$ and $4u_j^2$ yields $2u_1{u}_2{u}_3\le a^3/(3\sqrt{3})$. Thus its full volume $8u_1{u}_2{u}_3$ cannot exceed the value attained above; tilting cannot improve it. Boundary contact is allowed for the maximum; strict containment would give the same supremum.

Along a change in volume at fixed temperature, the first law of thermodynamics givesUse the Helmholtz Maxwell relation derived in (a) to replace the entropy derivative. This provesFor the stated Van der Waals equation, temperature-independent $a,b$ give $P_T{|}_V=R/(V-b)$. Hence $U_V{|}_T=a/V^2$. Integrating with respect to $V$ yieldsThe dilute-limit condition holds at every fixed $T$ and forces $h(T)=cT$, soThe result is the internal energy of a van der Waals gas in the fixed-amount convention of the question. Its volume derivative is positive even though the attractive energy correction itself is negative.