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For a continuously differentiable vector field on a bounded region with suitably piecewise smooth boundary, the divergence theorem giveswhere the normal vector points outward. In three dimensions, the position vector is $\mathbf{r}=(x,y,z)$ and $\nabla \cdot \mathbf{r}=1+1+1=3$. ThereforeThe result depends only on the enclosed volume, not on the shape or its location relative to the origin. Reversing the surface orientation reverses the sign.

Let $D$ be the disk $x^2+y^2\le 3$. The upper boundary is the graph $z=3-x^2-y^2$, with outward vector vector surface element $(2x,2y,1)dxdy$. Thus its flux isThe lower disk has outward normal vector $-{\mathbf{e}}_z$ and zero flux, since $F_z=0$. The total outward surface integral is therefore $9\pi$.

Independently, $\nabla \cdot \mathbf{F}=2$ and the enclosed volume isConsequently ${\iiint }_V\nabla \cdot \mathbf{F}dV=2V=\boxed{9\pi }$, agreeing with the divergence theorem. Both the orientation of the cap and inclusion of the closing disk have been accounted for.

For a general positive side length $s$, the cube has volume $s^3$, while separability of the integral givesThus the requested general dependence of the mean temperature isFor the cube with side length three this givesEquivalently each coordinate has volume-average value $s/2$, and the rectangular integration domain makes the mean of $xyz$ the product $(s/2{)}^3$.

A volume mean is the volume integral of the temperature divided by the volume. On this cube,Each one-dimensional integral is two, givingThe number is in the temperature units implicit in the given field law and coordinate units.

For a bounded volume $V$ with outward-oriented piecewise smooth boundary $S$, the divergence theorem statesHere $G$ is a continuously differentiable vector field on a neighborhood of $V$, $n$ the outward unit normal vector, and $dV$ the volume element. The left side is its outward flux.

For a twice continuously differentiable field $F$, equality of mixed partial derivatives givesApply the divergence theorem to $G=\nabla \times F$ to obtainThe regularity hypotheses justify differentiating and integrating the curl; a field singularity cannot simply be ignored in a volume argument.

For a continuously differentiable vector field $\mathbf{F}$ on a neighborhood of an oriented, piecewise smooth surface $S$, Stokes theorem statesHere $\partial S$ is the boundary curve, $d\mathbf{r}$ is its directed tangent element, $\hat{\mathbf{n}}$ is the chosen unit normal vector, $dS$ is the surface area element, and $\nabla \times \mathbf{F}$ is the curl. The right-hand rule relates the boundary direction to $\hat{\mathbf{n}}$. Thus circulation around the edge equals the flux of curl through the spanning surface. Reversing both orientations changes both signs. One can replace a difficult line integral by an easier surface integral, or replace one spanning surface by another within the region of regularity.

For a bounded volume $V$ with piecewise smooth closed boundary and a continuously differentiable vector field on a neighborhood of its closure, the divergence theorem statesHere ${\hat{\mathbf{n}}}_{\mathrm{out}}$ points out of $V$, including into any holes, $dV$ is the volume element, and $\nabla \cdot \mathbf{F}$ is the divergence. It equates total outward flux to integrated source strength. It converts a closed surface integral into a volume integral, and explains why internal fluxes cancel in a subdivided region. Singularities cannot be included without checking the hypotheses.

For a conservative force, the work around every closed path is zero. Stokes theorem gives zero flux of curl through every sufficiently small patch. Dividing by its area and shrinking the patch shows that every component of the continuous curl vanishes:Here $U$ is potential energy; conversely, zero curl implies a conservative vector field on a simply connected domain. The divergence theorem then gives the compatible relation ${\iint }_{\partial V}\mathbf{F}\cdot \hat{\mathbf{n}}dS=-{\iiint }_V{\nabla }^{2}UdV$. There is no general zero-divergence condition on a conservative force: $\mathbf{F}=-\mathbf{r}=-\nabla (r^2/2)$ is a concrete counterexample, with $\nabla \cdot \mathbf{F}=-3$.

Choose the normal pointing away from the origin, equivalently the outward normal of the first-octant tetrahedron. Parametrize the face by $\mathbf{r}=(x,y,2-x/2-y/2)$ on $D=\{x,y\ge 0:x+y\le 4\}$. Its oriented area element isThus the surface integral isThe divergence theorem checks this result: $\nabla \cdot \mathbf{F}=1$, the tetrahedron's volume is $4\cdot 4\cdot 2/6=16/3$, and each coordinate-plane face has zero flux. The opposite orientation gives $-16/3$; a signed flux requires an orientation.

With constant lake surface area, raising its level by $dh$ changes its volume by $S_{L}dh$. The symbol $M_L$ in the proposed balance is an effective filling rate measured as lake-level rise per unit time, so the volume inflow is $S_L{M}_L$. Volume conservation givesThus the lake rises when its inflow exceeds channel discharge and falls in the opposite case. If an inflow were instead specified directly as a volume rate $I_L$, its contribution would be $I_L/S_L$; keeping these units distinct is necessary for the interpretation of $M_L$.

For Röthlisberger channel drainage, $S$ is the water-filled cross-sectional area of the subglacial channel, $Q$ its volume discharge, $m$ the mass of ice melted per unit channel length per unit time, and $N=p_i-p_w$ the effective pressure, namely ice overburden pressure minus water pressure. The first balance converts melt mass to area growth through division by ice density ${\rho }_i$, and subtracts material creep closure $KS{N}^3$. The cubic pressure dependence comes from the adopted ice-flow law; $K$ includes its material and geometric factors.

The second relation balances the downslope hydraulic driving gradient ${\rho }_{w}g\sin \alpha$ against the adopted frictional pressure loss. It is an instantaneous discharge-area law, giving $Q=(\sin \alpha /f{)}^{1/2}{S}^{4/3}$ for positive downstream flow. The final balance equates latent heat $mL$ needed to melt the channel walls with the gravitational power dissipated per unit length, ${\rho }_{w}g\sin \alpha Q$. It assumes that this dissipated power is available for melting rather than substantial sensible heating or other energy losses. The model thus couples melt opening, pressure-driven material creep closure and hydraulic discharge.

The cross-sectional area $A(s,t)$ multiplied by the mean downstream speed $u(s,t)$ is the volume flux $Q=Au$. The term $A_t$ is local storage change per unit river length, $(Au{)}_s$ is net downstream export per unit length, and $M>0$ is lateral volume input per unit length per unit time, for example distributed tributary or runoff supply.

For a fixed reach $[s_1,s_2]$, incompressibility and constant density giveApply the fundamental theorem of calculus and localize the resulting integral identity to obtain $A_t+Q_s=M$. The density cancels because every term transports the same water; no sediment source is included in this balance.

Here $h$ is water depth, $u$ the depth-averaged downstream speed, $s$ bed elevation relative to the reference slope, and $\eta =s+h$ free-surface elevation relative to that reference. The constants ${\rho }_w,{\rho }_s$ are water and grain densities, $S$ is the reference bed slope, $\tau$ bed shear stress, $\varphi$ bed porosity, $c$ suspended sediment mass concentration, $q_b$ bedload solid volume flux per unit width, and $v_E,v_D$ erosion and deposition velocities. The settling speed is denoted by $v_s$ in the constitutive laws.

Depth integration of incompressible water conservation gives the depth-storage plus discharge-divergence balance. Depth-averaged momentum has local and advective acceleration on the left, and gravity driven by the actual surface gradient together with bed drag on the right. Hydrostatic pressure and a shallow-water approximation underlie this Saint-Venant description.

Suspended sediment conservation first has the conservative form $(hc{)}_t+(huc{)}_x={\rho }_s(v_E-v_D)$; the water balance reduces it to the material-derivative form printed in the source. Solid bed volume per unit horizontal area is $(1-\varphi )s$. Its storage plus bedload divergence equals deposition minus erosion, giving the Exner bed balance. Adding the suspended and bed balances after converting bed volume to grain mass cancels exchange terms: erosion and deposition transfer sediment between the two reservoirs rather than create it.

The cytoplasmic concentration $c$ gains a constant influx $r$ from outside the cell and loses calcium through a first-order extrusion term $kc$. The pump flux $J_{+}$ moves calcium from cytoplasm into the sarcoplasmic reticulum, saturating with cytoplasmic concentration through a Hill function with scale $K_1$, maximal rate $V_1$ and exponent $n$. The release flux $J_{-}$ returns stored calcium to the cytoplasm. It requires both calcium in the store and calcium-dependent activation of release; its two saturating factors have scales $K_2,K_3$ and exponents $m,p$. This represents calcium-induced calcium release. The leak $k_s{c}_s$ is passive return from the store.

Every internal transfer appears with opposite signs in the two balances, so $d(c+c_s)/dt=r-kc$ in the model's compatible volume normalization. Only influx and extrusion change the total. The nonlinear pump and activated release can redistribute this total rapidly and generate oscillations even with constant external input.

Take $a,b,c>0$ as the semiaxis lengths. For a centered axis-aligned box with positive half-edge lengths $x,y,z$, the volume is $V=8xyz$. Maximize $xyz$ under $g=x^2/a^2+y^2/b^2+z^2/c^2=1$. The Lagrange multiplier equations areMultiplying by $x,y,z$, respectively, shows that $x^2/a^2=y^2/b^2=z^2/c^2$. The constraint makes each value $1/3$. All boundary boxes with a zero edge have zero volume, while the positive stationary box exists. Compactness therefore gives the maximum box volume in an ellipsoidAllowing a displaced or differently oriented rectangular box cannot improve this result. Let $A=\mathrm{diag}(a,b,c)$, let $c_0$ be its center, and let $u_1,u_2,u_3$ be its perpendicular half-edge vectors. Averaging $|A^{-1}(c_0+\sum {\epsilon }_i{u}_i){|}^2\le 1$ over the eight signs gives $|A^{-1}{c}_0{|}^2+\sum |A^{-1}{u}_i{|}^2\le 1$. The determinant bound $|\det (A^{-1}{u}_1,A^{-1}{u}_2,A^{-1}{u}_3)|\le \prod |A^{-1}{u}_i|$, followed by the arithmetic-geometric mean inequality, makes its volume at most $8abc/(3\sqrt{3})$. The box found above attains the bound.

Represent the raindrops by a Poisson random measure on the space of location, impact time and volume. Physical volume is nonnegative, so take the intensity density to vanish for $v<0$, or equivalently work on ${\mathbb{R}}^2\times \mathbb{R}\times [0,\infty )$ and extend it by zero to ${\mathbb{R}}^4$. Assume the radius function is measurable and nonnegative for elapsed times at least zero. Its stated monotonicity means patches grow rather than dry away.

A drop contributes to wetting $\xi$ at time $\tau$ precisely if its impact time is no later than $\tau$ and its current patch contains $\xi$. Thus define the deterministic measurable setThe number of covering drops is $\Pi (A_{\xi ,\tau })$. Its intensity measure of a point process isSince the point is dry if and only if that count is zero, the Poisson count definition gives the void probability for growing Poisson rain patchesHere $e^{-\infty }=0$: when the intensity is infinite, exhaust the covering set by increasing finite-intensity sets; their decreasing zero-count events have probabilities tending to zero. This also covers storms with infinitely many drops that can reach the point. The formula is valid for spatially inhomogeneous intensity. Only if the intensity is spatially constant may the spatial integral be replaced by $\pi r(\tau -t,v{)}^2$ times that intensity. Whether the circular boundary is included makes no difference for a density with respect to Lebesgue measure, since each such boundary has zero spatial area. This construction is a time-dependent Poisson germ-grain model.

For total rainfall, the required general result is the Campbell first-moment formula: if $\Pi$ is a Poisson random measure with intensity $\mu$ and $h$ is any nonnegative measurable function, thenwith infinite values permitted. For a signed $h$, it holds with finite absolute expectation whenever $\int |h|d\mu <\infty$. No square-integrability hypothesis is required for this first-moment identity. The result is stated rather than proved, as allowed here.

The total rainfall volume is $V_{\mathrm{total}}={\sum }_{(X,T,V)\in \Pi }V=\int v\Pi (dxdtdv)$, counting every drop once, independently of overlaps between its patch and other patches. Apply the stated formula with $h(x,t,v)=v$ on the physical mark space. Extending by zero to negative volume givesIf the displayed nonnegative integral is finite, the total volume is integrable and finite almost surely. If it is infinite, its expectation is infinite; this alone does not imply that the random total is infinite almost surely. The radius affects wet-area coverage but does not change this total-volume expectation.

In the thin-current approximation, vertical momentum balance is hydrostatic: $p=p_{\mathrm{atm}}+\rho g(h-z)$. Neglect inertia and radial viscous derivatives compared with the vertical ones. The radial Stokes equation is $\mu u_{zz}=p_r=\rho g{h}_r$, with no slip at $z=0$ and zero tangential stress $u_z=0$ at $z=h$. Integration givesIts depth-integrated radial flux is $Q_r={\int }_0^{h}udz=-\rho g{h}^3{h}_r/(3\mu )$. Axisymmetric mass conservation $h_t+r^{-1}{\partial }_r(r{Q}_r)=0$ therefore givesThe remaining conditions are conserved volume $2\pi {\int }_0^{R(t)}rhdr=V$, regularity and zero radial flux at the center, $h(R(t),t)=0$, zero thickness outside the front, and the initially concentrated volume at the origin. The advancing edge obeys $R^{\prime }={\lim }_{r\uparrow R}{Q}_r/h$ when this limit exists. One must not impose a finite zero slope at this compact-support edge.

Put $h=t^{a}F(r{t}^{-b})$ with dimensional factors to be restored. Constant volume gives $a+2b=0$, while balancing the equation gives $a-1=4a-2b$. Hence $a=-1/4,b=1/8$. In normalized similarity variables $\eta =r/(\beta V^{3}t{)}^{1/8}$ and $h=V^{1/4}(\beta t{)}^{-1/4}F(\eta )$, the equation becomesMultiplying by $\eta$ and integrating from the regular center yields $\eta F^3{F}^{\prime }=-{\eta }^{2}F/8$. Where $F>0$, this integrates to $F^3=3({\eta }_0^2-{\eta }^2)/16$. Restoring physical variables giveswith zero thickness outside. Write $h=A(t)(1-r^2/R^2{)}^{1/3}$. Its volume is $3\pi A{R}^2/4$, so $A=4V/(3\pi R^2)$. Combining this with $A^3=3R^2/(16\beta t)$ givesThe center height scales as $t^{-1/4}$, while the front scales as $t^{1/8}$. The ideal profile has a singular slope at the edge; microscopic contact-line physics is excluded from the lubrication model, which applies to the sufficiently spread bulk current.

Take the divergence of the Ampere-Maxwell law and use Gauss's law $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. Since the divergence of a curl is zero, this gives the charge continuity equation ${\partial }_t\rho +\nabla \cdot \mathbf{j}=0$. Uniform conductivity and Ohm's law imply $\nabla \cdot \mathbf{j}=\sigma \nabla \cdot \mathbf{E}=\sigma \rho /{ϵ}_0$. ThereforeThe spatially uniform initial density stays uniform in the interior; $\tau$ is the charge relaxation time.

Denote the cylinder radius by $R$. Cylindrical symmetry makes the interior electric field radial. Gauss's law on a coaxial cylinder of radius $r<R$ and length $\ell$ gives $2\pi r\ell E_r=\rho (t)\pi r^2\ell /{ϵ}_0$. ThusThe current does not disappear at the interface: the vacuum conducts no current, so charge accumulates on the cylindrical surface. Initially taking the charge to be the stated volume charge alone, its surface charge density satisfies ${\dot{\rho }}_s=j_r(R^{-},t)$, giving the charge relaxation in an isolated conducting cylinder lawThe total enclosed charge per unit length is $\pi R^2\rho (t)+2\pi R{\rho }_s(t)=\pi R^2{\rho }_0$. Consequently the exterior electric field isindependent of time. The boundary jump obeys $E_r(R^{+})-E_r(R^{-})={\rho }_s/{ϵ}_0$, as it must. Applying Gauss's law to the decaying volume charge alone would incorrectly predict a decaying exterior electric field and violate electric charge conservation. The vector formulas also specify the inward direction if ${\rho }_0<0$.

There are two necessary qualifications to the literal formulation. The whole group $G$ is itself a closed normal subgroup, so the claim about containment in the centre must concern proper closed normal subgroups. Also, the homomorphism conclusion needs a nontrivial target $H$: for example, $SU(2)\to \{1\}$ has surjective differential onto the zero Lie algebra, but its group kernel is all of $SU(2)$ and no quotient by a finite subgroup is a one-point group. We prove the precise assertions, including these exceptional cases.

Write $\mathfrak{g}=\mathrm{Lie}(G)$. The centre $\mathfrak{z}(\mathfrak{g})$ is invariant under the Adjoint representation, so irreducibility makes it either zero or all of $\mathfrak{g}$. In the latter case $\mathfrak{g}$ is Abelian. For connected $G$, the adjoint action is then trivial: the group is generated by exponentials, and ${\mathrm{Ad}}_{\exp X}=e^{\mathrm{ad}X}=I$. A nonzero trivial representation is irreducible only in dimension one. A connected compact one-dimensional Lie group is a circle, so $G\cong U(1)$.

For every other $G$, the Lie algebra of its centre is zero. The centre is a closed compact Lie subgroup; a zero-dimensional compact Lie group is finite. Let $N$ be a closed normal subgroup. The closed-subgroup theorem identifies $\mathfrak{n}=\mathrm{Lie}(N)$ as an adjoint-invariant subspace of $\mathfrak{g}$, hence $\mathfrak{n}=0$ or $\mathfrak{n}=\mathfrak{g}$. If the latter holds, $N$ contains an identity neighbourhood and is open, so connectedness forces $N=G$. If $N$ is proper, it therefore has zero Lie algebra and is finite. For any $a\in N$, the continuous map $g\mapsto ga{g}^{-1}$ takes connected $G$ into the finite set $N$, so it is constant and equal to $a$. ThusProper closed subgroups of the circle are finite as well, and of course central.

Now let $d\Phi$ be surjective. The submersion theorem makes $\Phi (G)$ contain an identity neighbourhood in $H$, so it is an open subgroup. Since $H$ is connected, $\Phi (G)=H$. Its group kernel $F$ is closed and normal. If $H$ is nontrivial, $F$ is proper and the preceding argument gives a finite central group kernel; this also holds for $G\cong U(1)$, whose proper closed subgroups are finite. The Lie-group first isomorphism theorem givesIf $H$ is trivial, the correct statement is instead $H=G/G$.

For the Spin groups, use the real Clifford algebra with $v^2=-\Vert v{\Vert }^2$. The group $\mathrm{Spin}(m)$ consists of products of an even number of unit vectors inside its invertible even part. Conjugation on the embedded ${\mathbb{R}}^m$ defines the standard double covering $\mathrm{Spin}(m)\to SO(m)$ with group kernel $\{\pm 1\}$. The Clifford relations show that a unit-vector reflection, with the usual twisted conjugation, is a hyperplane reflection; the even products give orientation-preserving transformations. The Cartan–Dieudonné theorem states that every orthogonal transformation is a product of reflections, proving surjectivity. For $m\ge 3$ this connected covering group is simply connected; $\mathrm{Spin}(2)$ is a circle.

Here is an explicit exterior-square construction of the requested six-dimensional map. Put $E={\mathbb{C}}^4$ with its standard Hermitian form and volume $e_1\wedge e_2\wedge e_3\wedge e_4$. On ${\Lambda }^{2}E$, write $e_{ij}=e_i\wedge e_j$ and define an antilinear map $J$ byIt satisfies $J^2=1$. Intrinsically, if $u\wedge v=B(u,v)\mathrm{vol}$ and the Hermitian inner product $h$ is linear in its first variable, then $B(u,Jv)=h(u,v)$. Because $SU(4)$ preserves both $B$ and $h$, it commutes with $J$. This is the real structure of the SU4 exterior square.

The fixed space $W$ has the following real orthonormal basis, with all vectors divided by $\sqrt{2}$:Thus $W$ is six-dimensional over $\mathbb{R}$, its complexification is ${\Lambda }^{2}E$, and the exterior-square action restricts to an orthogonal representation on $W$. Its determinant is one because $SU(4)$ is connected. We obtain $R:SU(4)\to SO(W)\cong SO(6)$.

To compute the group kernel, diagonalize $g\in SU(4)$ with eigenvalues $z_1,\dots ,z_4$. If $R(g)=I$, its complexified exterior-square action is the identity, hence $z_i{z}_j=1$ for every $i<j$. Comparing pairs shows that all $z_i$ are equal, and their common value has square one. A unitary matrix with that single eigenvalue is scalar, so $\ker R=\{\pm I_4\}$. Conversely, those two scalar matrices plainly act trivially on the exterior square.

The differential of $R$ is injective because its group kernel is finite. Both Lie algebras have dimension $15$, so the differential is an isomorphism. Its image is consequently open in connected $SO(6)$, hence all of $SO(6)$. ThereforeThis also realizes $SU(4)$ as $\mathrm{Spin}(6)$, since $SU(4)$ is simply connected, by uniqueness of the connected simply connected covering group.

A local stellar relaxation time depends on the number density $n$ of scatterers. Calling $N$ a total number, as the PDF does, leaves the displayed expression dimensionally wrong: ${\sigma }^3/(G^2{m}^{2}N)$ has units of time per volume. The intended local expression must use $n$, or $N/V$ for a homogeneous sample of volume $V$. We shall also make the normalization of the stellar relaxation time explicit, since an unspecified representative one-dimensional velocity dispersion does not fix an exact numerical coefficient.

In a small-angle gravitational encounter, approximate the relative trajectory by a straight line of speed $u$ and impact parameter $b$. The perpendicular acceleration of one star due to the other is $Gmb/(b^2+u^2{t}^2{)}^{3/2}$. Its impulse is thereforeA star encounters $2\pi bdbnu$ scatterers per unit time in this interval of impact parameters. Independent deflections have random directions, so their variances add as in a random walk. Hence the accumulated squared velocity change has rateThe Coulomb logarithm in stellar dynamics arises because each logarithmic interval of impact parameter contributes equally. The lower cutoff is approximately the strong-deflection scale, $b_{\min }\simeq G(2m)/u^2$ for equal masses, or the physical collision scale if larger. The upper cutoff is a scale on which the local density and straight-line encounter approximation cease to apply, such as the local disk thickness. Thus $\Lambda$ is dimensionless; it is not a number density or an arbitrary dimensional argument of a logarithm.

To display the normalization issue, use the representative relative speed $u=\sqrt{2}\sigma$ and define $T(\kappa )$ by $D_{v}T=\kappa {\sigma }^2$. This givesChoosing $\kappa =3/4$ recovers precisely the printed coefficient after repairing $N$ to $n$:This is an explicitly defined convention, not a uniquely derivable normalization from the supplied data. If instead relaxation means a mean-square change equal to the full initial random-speed variance, $3{\sigma }^2$, the same encounter approximation gives $T_R=4T_R^{\mathrm{printed}\mathrm{convention}}$. Averaging the encounter speed distribution instead of using one representative $u$ changes the coefficient again. The robust result is the scaling $T_R\propto {\sigma }^3/(n{G}^2{m}^2\ln \Lambda )$.

For a transparent solar-neighbourhood estimate, take $n=0.1{\mathrm{pc}}^{-3}$, $m=M_{\odot }$, $\sigma =30{\mathrm{km}\mathrm{s}}^{-1}$ and $b_{\max }=300\mathrm{pc}$. With $G=4.3009\times 10^{-3}\mathrm{pc}{M}_{\odot }^{-1}({\mathrm{km}\mathrm{s}}^{-1}{)}^2$, one obtains $b_{\min }=4.78\times 10^{-6}\mathrm{pc}$, $\ln \Lambda =17.96$, andThese illustrative parameters give an order $10^{13}$--$10^{14}$ year stellar relaxation time, much longer than a galactic lifetime of order $10^{10}$ years. The solar neighbourhood is consequently a collisionless stellar system with respect to encounters between individual stars. Scattering by massive clouds and collective structure is a different process; the estimate does not assert that every mechanism of disk heating is negligible.

The vector triple product removes the nested cross products:Since $\mathbf{\omega }$ is constant, the divergence isThe cube has volume $2^3=8$. With the outward unit normal, the divergence theorem therefore gives

For a regular smooth constraint $g=0$, meaning $\nabla g\ne 0$ at the point, constrained stationarity requires $\nabla f=\lambda \nabla g$. Solve these Lagrange multiplier equations together with the constraint, then compare candidate values and possible boundary or singular cases.

Let $r>0$ and $h>0$ be cylinder radius and height. Its volume is $V=\pi r^{2}h$ and the surface-area constraint is $2\pi r^2+2\pi rh=A$. The multiplier equations areThe second gives $\lambda =r/2$, and substituting into the first gives $h=2r$. Consequently $6\pi r^2=A$. To check global maximality, eliminate $h$ and obtain $V(r)=Ar/2-\pi r^3$ on $0<r<\sqrt{A/(2\pi )}$; it tends to zero at both ends and has exactly this one stationary point. The largest volume is