The Divergence Theorem and Classical Stokes: Examples and Counterexamples

Each of the six maps is affine, so its two parameter derivatives are the constant standard basis vectors ${\phi }_{z^{+},u}=e_x$, ${\phi }_{z^{+},v}=e_y$; ${\phi }_{z^{-},u}=e_y$, ${\phi }_{z^{-},v}=e_x$; ${\phi }_{x^{+},u}=e_y$, ${\phi }_{x^{+},v}=e_z$; ${\phi }_{x^{-},u}=e_z$, ${\phi }_{x^{-},v}=e_y$; ${\phi }_{y^{+},u}=e_z$, ${\phi }_{y^{+},v}=e_x$; ${\phi }_{y^{-},u}=e_x$, ${\phi }_{y^{-},v}=e_z$. Computing each cross product from [F1] gives $e_x\times e_y=(0,0,1)=e_z$, $e_y\times e_x=-e_z$, $e_y\times e_z=(1,0,0)=e_x$, $e_z\times e_y=-e_x$, $e_z\times e_x=(0,1,0)=e_y$ and $e_x\times e_z=-e_y$, so the six oriented area vectors are the constants $e_z,-e_z,e_x,-e_x,e_y,-e_y$ as displayed.

The three quadruples $(z,[0,1{]}^2,0,1)$, $(x,[0,1{]}^2,0,1)$ and $(y,[0,1{]}^2,0,1)$, with constant graph functions, are simple descriptions of $B$ in the three directions in the sense of [F4]: the base $[0,1{]}^2$ is compact, Jordan measurable and has nonempty interior; the constants $0<1$ satisfy the weak and the strict inequality; and in each case $\{p:{\pi }_k(p)\in [0,1{]}^2,0\le p_k\le 1\}$ is exactly $[0,1{]}^3$, because ${\pi }_k$ lists the two coordinates other than the $k$th.

Each of the six pairs $(S,\phi )$ is a regular patch in the sense of [F2]: the square $S$ is compact and Jordan measurable, being a rectangle, and is the closure of its nonempty convex, hence connected, interior $(0,1{)}^2$; each $\phi$ is affine and therefore $C^1$ on all of ${\mathbb{R}}^2$; each oriented area vector is a nonzero constant by step 1.1; and each $\phi$ is injective on ${\mathbb{R}}^2$, since its two direction vectors are distinct standard basis vectors and reading the two matching coordinates of the image recovers $(u,v)$, so in particular no interior parameter point shares its image with a distinct point of $S$.

Take ${\mathrm{\Sigma }}_z^{+}=({\phi }_{z^{+}})$, ${\mathrm{\Sigma }}_z^{-}=({\phi }_{z^{-}})$ and ${\mathrm{\Sigma }}_z^0=({\phi }_{x^{+}},{\phi }_{x^{-}},{\phi }_{y^{+}},{\phi }_{y^{-}})$. The image of ${\phi }_{z^{+}}$ is the face $p_z=1$, the graph of ${\gamma }_2=1$, and by step 1.1 the $z$ coordinate of its oriented area vector is $1>0$; the image of ${\phi }_{z^{-}}$ is the graph of ${\gamma }_1=0$ with $z$ coordinate $-1<0$; and the four lateral vectors $\pm e_x,\pm e_y$ have $z$ coordinate $0$, so that coordinate vanishes on the whole parameter square. The projected image ${\pi }_z[{\phi }_{z^{+}}[(0,1{)}^2]]$ is $(0,1{)}^2$, whose complement in $[0,1{]}^2$ is $\mathrm{\partial }([0,1{]}^2)$ of content zero by [F8], and likewise for ${\phi }_{z^{-}}$, where ${\pi }_z(v,u,0)=(v,u)$; each graph sublist is a single patch, so pairwise disjointness is vacuous, and both are nonempty. So the presentation is adapted to the $z$ description of step 1.2, in the sense of [F5].

Take ${\mathrm{\Sigma }}_x^{+}=({\phi }_{x^{+}})$, ${\mathrm{\Sigma }}_x^{-}=({\phi }_{x^{-}})$ and ${\mathrm{\Sigma }}_x^0=({\phi }_{z^{+}},{\phi }_{z^{-}},{\phi }_{y^{+}},{\phi }_{y^{-}})$. By step 1.1 the $x$ coordinates of the oriented area vectors are $1$ for ${\phi }_{x^{+}}$, $-1$ for ${\phi }_{x^{-}}$ and $0$ for the other four. The image of ${\phi }_{x^{+}}$ is the face $p_x=1$ and ${\pi }_x(1,u,v)=(u,v)$, so its projected image is $(0,1{)}^2$; the image of ${\phi }_{x^{-}}$ is $p_x=0$ and ${\pi }_x(0,v,u)=(v,u)$, again with projected image $(0,1{)}^2$. Both complements in the base are $\mathrm{\partial }([0,1{]}^2)$, of content zero. So the same presentation is adapted to the $x$ description.

Take ${\mathrm{\Sigma }}_y^{+}=({\phi }_{y^{+}})$, ${\mathrm{\Sigma }}_y^{-}=({\phi }_{y^{-}})$ and ${\mathrm{\Sigma }}_y^0=({\phi }_{z^{+}},{\phi }_{z^{-}},{\phi }_{x^{+}},{\phi }_{x^{-}})$. By step 1.1 the $y$ coordinates of the oriented area vectors are $1$ for ${\phi }_{y^{+}}$, $-1$ for ${\phi }_{y^{-}}$ and $0$ for the other four. The image of ${\phi }_{y^{+}}$ is the face $p_y=1$ and ${\pi }_y(v,1,u)=(u,v)$, so its projected image is $(0,1{)}^2$; the image of ${\phi }_{y^{-}}$ is $p_y=0$ and ${\pi }_y(u,0,v)=(v,u)$, with projected image $(0,1{)}^2$. So the same presentation is adapted to the $y$ description.

The six images are the six closed faces of $B$, each contained in $\mathrm{\partial }B$, and their union is $\mathrm{\partial }B$ by [F7], since a point of $B$ fails to be interior exactly when one of its coordinates is $0$ or $1$. Two distinct faces meet in a closed edge, a vertex or the empty set; the preimage of such an intersection in either parameter square is contained in $\mathrm{\partial }S$, which has content zero by [F8], so the overlap condition of [F3] holds. Interior parameter points map into the six open faces, which are pairwise disjoint, so no point of an overlap is the image of an interior parameter point of two distinct patches and the normal-agreement condition of [F3] holds with nothing to check. Hence the six patches form a compatible finite patch presentation of $\mathrm{\partial }B$.

Steps 2.1 and 3.1 make the six patches a compatible finite patch presentation of $\mathrm{\partial }B$, step 1.2 supplies the three simple descriptions, and steps 2.2, 2.3 and 2.4 make that one presentation adapted in all three directions. By [F6] the box $B$, with these data, is an elementary solid region.

The three components of $F$ are $x^2$, $y^2$ and $z^2$, so by [L5] the partial derivatives ${\mathrm{\partial }}_x{F}_x=2x$, ${\mathrm{\partial }}_y{F}_y=2y$ and ${\mathrm{\partial }}_z{F}_z=2z$ exist and are continuous on ${\mathbb{R}}^3$, as are the six off-diagonal ones, which vanish. Hence $F$ is $C^1$ on ${\mathbb{R}}^3$ by [F1] and $\mathrm{div}F=2x+2y+2z=2(x+y+z)$.

By [L1] the box $B$ with those six patches is an elementary solid region, and ${\mathbb{R}}^3$ is an open set containing it, so [L2] applies to $F$ on $B$.

The function $2(x+y+z)$ is continuous on the compact Jordan set $B$, hence integrable by [L6]. Applying [L3] to split off the $z$ coordinate and then again to split off $y$, and evaluating each inner integral by [L4] and [L5]: ${\int }_0^{1}2(x+y+z)dz=2(x+y)+1$, then ${\int }_0^1(2(x+y)+1)dy=2x+2$, then ${\int }_0^1(2x+2)dx=3$. So ${\iiint }_B\mathrm{div}F=3$ by step 1.1.

By [F2] and [F3] the six face fluxes are the integrals over $S$ of $⟨F({\phi }_j),{\phi }_{j,u}\times {\phi }_{j,v}⟩$, which by [L1] is $\pm$ one coordinate of $F$ along the patch. Face by face: on ${\phi }_{z^{+}}$ it is $F_z(u,v,1)=1$; on ${\phi }_{z^{-}}$ it is $-F_z(v,u,0)=0$; on ${\phi }_{x^{+}}$ it is $F_x(1,u,v)=1$; on ${\phi }_{x^{-}}$ it is $-F_x(0,v,u)=0$; on ${\phi }_{y^{+}}$ it is $F_y(v,1,u)=1$; and on ${\phi }_{y^{-}}$ it is $-F_y(u,0,v)=0$. Since $S=[0,1{]}^2$, [L3], [L4], and [L5] give ${\int }_{S}1={\int }_0^1{\int }_0^{1}1dudv=1$, while ${\int }_{S}0=0$; so the six patch fluxes are $1,0,1,0,1,0$ and the boundary flux is their sum, $3$.

Steps 2.1 and 2.2 give $3$ for the volume integral and $3$ for the boundary flux, which is what [L2] asserts of them.

The eight parameter rectangles are $I_{\pm }\times J_j$, obtained by cutting the spherical parameter rectangle at the quarter turns named in [L5].

Differentiating gives ${\phi }_{\varphi }=R(\cos \varphi \cos \theta ,\cos \varphi \sin \theta ,-\sin \varphi )$ and ${\phi }_{\theta }=R(-\sin \varphi \sin \theta ,\sin \varphi \cos \theta ,0)$, and [F6], [L2], and [L3] give ${\phi }_{\varphi }\times {\phi }_{\theta }=R\sin \varphi \phi (\varphi ,\theta )$.

On the interior of each rectangle one has $\sin \varphi >0$ by [L4], so step 2.1 gives a nonzero oriented area vector. If two interior parameter points have the same image, their third coordinates give the same $\cos \varphi$; strict monotonicity of cosine on $(0,\pi )$ gives the same $\varphi$, and the first two coordinates then give the same point of the unit circle, so [L9] gives the same $\theta$. Thus each restriction is injective on its interior. Two distinct octants meet only along boundary arcs whose preimages have content zero and contain no point that is interior for both patches. Hence the eight restrictions are regular and compatible in the senses of [F4] and [F5].

In the $z$ direction the base is the closed disc $D_z=\{(x,y):x^2+y^2\le R^2\}$ and the two boundary functions are ${\gamma }_1^z(x,y)=-\sqrt{R^2-x^2-y^2}$ and ${\gamma }_2^z(x,y)=\sqrt{R^2-x^2-y^2}$, so [F2] describes $B_R$; the third coordinate of the oriented area vector is $R^2\sin \varphi \cos \varphi$, so by [L4] the four patches with $\varphi \in (0,\pi \mathrm{/}2)$ form the upper sublist and the four with $\varphi \in (\pi \mathrm{/}2,\pi )$ form the lower sublist, with no lateral patch because the vanishing set $\sin \varphi =0$ or $\cos \varphi =0$ lies on parameter boundaries.

In the $x$ direction the base is the closed disc $D_x=\{(y,z):y^2+z^2\le R^2\}$ and the boundary functions are ${\gamma }_1^x(y,z)=-\sqrt{R^2-y^2-z^2}$ and ${\gamma }_2^x(y,z)=\sqrt{R^2-y^2-z^2}$, so [F2] again describes $B_R$. The first coordinate of the oriented area vector is $R^2{\sin }^2\varphi \cos \theta$, so by [L4] the four octants with $\theta \in (0,\pi \mathrm{/}2)\cup (3\pi \mathrm{/}2,2\pi )$ form the upper sublist and the other four the lower sublist; the vanishing set $\sin \varphi =0$ or $\cos \theta =0$ lies on parameter boundaries, so there is no lateral patch.

In the $y$ direction the base is the closed disc $D_y=\{(z,x):z^2+x^2\le R^2\}$ and the boundary functions are ${\gamma }_1^y(z,x)=-\sqrt{R^2-z^2-x^2}$ and ${\gamma }_2^y(z,x)=\sqrt{R^2-z^2-x^2}$, so [F2] describes $B_R$ a third time. The second coordinate of the oriented area vector is $R^2{\sin }^2\varphi \sin \theta$, so the split is by $\theta \in (0,\pi )$ against $\theta \in (\pi ,2\pi )$; again the vanishing set lies on parameter boundaries, so there is no lateral patch.

The projections of the interiors of the four upper octants onto the $xy$ plane are the four open quarter discs, pairwise disjoint, and the same is true for the four lower octants; each union misses only the two coordinate diameters and the boundary circle of $D_z$. The circle has content zero because the closed disc is Jordan measurable by [L6] and [L7], so its boundary has content zero; each diameter is a continuous graph over a compact interval and has content zero by [L8]; and the finite union of those three sets has content zero by [F9]. Thus both graph sublists satisfy the coverage clause in the $z$ direction.

In the $x$ direction the projections onto the $yz$ plane of the four upper octants are the four open quarter discs of $D_x$, pairwise disjoint: $J_0$ gives the half with $y>0$ and $J_3$ the half with $y<0$, and in each half the two choices $\varphi \in (0,\pi \mathrm{/}2)$ and $\varphi \in (\pi \mathrm{/}2,\pi )$ split by the sign of $z$. The four lower octants have the same projected images, now coming from $J_1$ and $J_2$. In each case the omitted set is the union of the two coordinate diameters and the boundary circle of $D_x$, which has content zero by the same argument as in step 4.1. Thus both graph sublists satisfy the coverage clause in the $x$ direction.

In the $y$ direction the projections onto the $zx$ plane of the four upper octants are the four open quarter discs of $D_y$, pairwise disjoint: $\theta \in (0,\pi )$ gives the half with $x>0$ or $x<0$ according to whether $\theta \in (0,\pi \mathrm{/}2)$ or $(\pi \mathrm{/}2,\pi )$, and the two halves are split again by the sign of $z$. The four lower octants have the same projected images. The omitted set is the union of the two coordinate diameters and the boundary circle of $D_y$, hence has content zero by the same argument as in step 4.1. So both graph sublists satisfy the coverage clause in the $y$ direction.

Steps 3.1, 3.2, 4.1, 3.3, 5.1, 3.4, and 5.2 show that the same eight patches are compatible and adapted in all three coordinate directions, and step 2.1 gives them the outward orientation. Therefore [F1] makes $B_R$ with this presentation an elementary solid region.

On each spherical octant patch, [L1], [F2], [F4], and [L7] give $⟨X(\phi (\varphi ,\theta )),{\phi }_{\varphi }\times {\phi }_{\theta }⟩=⟨\phi (\varphi ,\theta ),R\sin \varphi \phi (\varphi ,\theta )⟩=R^3\sin \varphi$.

On each of the four upper octant patches, step 1.1 gives the flux integrand $R^3\sin \varphi$ on a parameter rectangle $[0,\pi \mathrm{/}2]\times [\alpha ,\alpha +\pi \mathrm{/}2]$; for fixed $\varphi$ the $\theta$-section is constant, and for fixed $\theta$ the $\varphi$-section is $R^3\sin \varphi$, so [L4], [L5], and [L6] give the flux value $R^3{\int }_{\alpha }^{\alpha +\pi \mathrm{/}2}{\int }_0^{\pi \mathrm{/}2}\sin \varphi d\varphi d\theta =R^3(\pi \mathrm{/}2)$. On each of the four lower octant patches the same argument gives $R^3{\int }_{\alpha }^{\alpha +\pi \mathrm{/}2}{\int }_{\pi \mathrm{/}2}^{\pi }\sin \varphi d\varphi d\theta =R^3(\pi \mathrm{/}2)$. Summing the eight patch fluxes by [F3] yields the total outward flux $8\cdot R^3(\pi \mathrm{/}2)=4\pi R^3$.

The field $X$ has divergence $3$ by [F1], so [L3] and [L2] both identify the content of $B_R$ with one third of the flux computed in step 2.1, namely $(4\mathrm{/}3)\pi R^3$.

This agrees with the published volume formula [L8]. An inward presentation would reverse the sign of the flux, so the agreement is a check on the orientation convention as well as a computation of the volume.

The two caps and four quarter-cylinder side patches are the six displayed parametrizations, with the quarter-turn cuts in the azimuth chosen as in [L5].

Differentiating gives ${\kappa }_{+,r}\times {\kappa }_{+,\theta }=(0,0,r)$, ${\kappa }_{-,r}\times {\kappa }_{-,\theta }=(0,0,-r)$, and ${\sigma }_{j,\theta }\times {\sigma }_{j,z}=R(\cos \theta ,\sin \theta ,0)$, so the caps have oriented area vectors $\pm r{e}_z$ and each side quarter has outward horizontal area vector $R(\cos \theta ,\sin \theta ,0)$ by [F6], [L2], and [L3].