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If $X$ is an oriented $n$-manifold with boundary, $F:\partial X\hookrightarrow X$ is the inclusion, and $\omega$ is a compactly supported $(n-1)$-form, thenwhere the boundary has the outward-normal-first boundary orientation. A subordinate partition of unity reduces the theorem to the fundamental theorem of calculus in oriented coordinate half-spaces; local finiteness and compact support ensure that only finitely many terms contribute.

For the counterclockwise unit-square boundary and $\omega =xdy$, the right-hand side is $1$, while the left-hand side is ${\int }_{[0,1{]}^2}dx\wedge dy=1$. The equation says that the boundary integral measures the total local circulation in the region. Reversing the orientation of a smooth manifold reverses both sides together.

In dimension one, the boundary of an interval consists of its endpoint with a plus sign and its starting point with a minus sign; the identity becomes ${\int }_a^{b}df=f(b)-f(a)$. In the plane it gives Green theorem. Applied to appropriate differential forms in three dimensions it gives both Stokes theorem and the divergence theorem. The boundary and the differential being integrated must belong to the stated domain: removing a singular point can make a spanning disc unavailable.

Orientability asks whether local choices of handedness can agree everywhere on a manifold. On a smooth manifold, an orientation of a smooth manifold chooses an orientation of a vector space for each tangent space, continuously with position. An ordered basis describes the same handedness as another one when the change-of-basis determinant is positive. An oriented atlas implements this condition through positive Jacobian determinants of the smooth transition maps.

A paper strip makes the obstruction visible. Join its short ends without twisting: the resulting untwisted band admits a consistent choice of unit normal. Add a half-turn before joining: on the Möbius band, following the core circle of a Möbius band reverses that choice when you first return to the same point. A local decision about handedness has encountered a global contradiction. The normal reversal around the Möbius core makes this precise for an explicit model.

Both rectangles have coordinates $(u,v)\in [0,1]\times [-1,1]$. The untwisted band identifies $(0,v)$ with $(1,v)$; the Möbius band identifies $(0,v)$ with $(1,-v)$. Across a seam the corresponding local coordinate changes preserve or reverse $v$, giving determinants $+1$ and $-1$. This sign explains the handedness change; the drawing also tracks the two boundary circles of the untwisted band and the single boundary circle of the Möbius band.

For a smooth topological surface embedded in ${\mathbb{R}}^3$, choosing an orientation of a surface is equivalent to choosing a continuous unit normal. This picture is specific to surfaces in three-dimensional space: the general definition uses compatible local orientations, without requiring an embedding. Orientability is a topological invariant, so a homeomorphism cannot turn a Möbius band into a untwisted band. The sphere and torus are orientable surfaces; the Klein bottle is not. ^[ref]

The distinction matters when a sign has geometric meaning. A signed flux surface integral needs an orientation of a surface, and Stokes theorem relates that choice to the outward-normal-first boundary orientation. Ordinary unsigned surface area does not require orientability. Thus a Möbius band still has a perfectly meaningful area even though it has no continuous global unit normal.

A connected orientable manifold of positive dimension has two possible orientations: reversing every local choice gives the other one. If a manifold has several connected components, each component can be oriented independently when it is orientable. For topological surfaces, a useful obstruction criterion is the presence of a Möbius band: a surface is nonorientable exactly when it contains such a band. ^[ref]

A differential form of degree $k$ is integrated over an oriented $k$-dimensional smooth manifold by integrating its coefficient in orientation-preserving coordinates. For an oriented parametrization $F:U\to M$, this means integrating the pullback $F^{\ast }\alpha$ over $U$; a partition of unity combines charts. The change of variables formula makes the result independent of the charts. Reversing the orientation of a smooth manifold reverses the integral. The Generalized Stokes theorem says, for a compact oriented $k$-dimensional smooth manifold with boundary and a smooth $(k-1)$-form $\beta$,where the right-hand side includes the boundary pullback, and the boundary has the outward-normal-first boundary orientation. Compact support suffices on a noncompact manifold. This unifies the fundamental theorem of calculus, circulation and flux identities.

For example, take the unit disk $D$ oriented by $dx\wedge dy$ and the differential one-form $\beta =\frac{1}{2}(xdy-ydx)$. Its exterior derivative is $d\beta =dx\wedge dy$, so the area integral is $\pi$. Parametrizing the positively oriented circle by $(x,y)=(\cos t,\sin t)$ gives the pullback $\beta =\frac{1}{2}dt$ and the boundary integral ${\int }_0^{2\pi }\frac{1}{2}dt=\pi$, directly illustrating Generalized Stokes theorem.

For the distance calculation, write $u=|{\mathbf{r}}_1|$, $v=|{\mathbf{r}}_2|$, $x=u+v$ and $y=u-v$; these are scalar radii, not position vectors. The triangle inequality and its reverse give $r\le u+v=x$ and $|y|=|u-v|\le r$. Also,Thus $r\le x\le 2R-|y|$ and $|y|\le r$. These conditions, with $u,v\ge 0$, are also sufficient for a triangle with side lengths $u,v,r$. Endpoints correspond to collinear configurations.

Assume the two positions are independent random variables, each having constant volume probability density inside the radius-$R$ ball. Independence is needed: uniform marginal distributions alone do not determine the distance distribution. Put $C=3/(4\pi R^3)$ and $\mathbf{s}={\mathbf{r}}_2-{\mathbf{r}}_1$. The change of variables $({\mathbf{r}}_1,{\mathbf{r}}_2)\mapsto ({\mathbf{r}}_1,\mathbf{s})$ has unit Jacobian determinant, so the joint probability volume differential form is the product of the two normalized volume forms. In spherical coordinates for ${\mathbf{r}}_1$ and for $\mathbf{s}$ relative to its axis, it isThe polar coordinates fail on axes and at zero radii, which are sets of zero volume and do not affect the probability. Here $\theta$ is the angle from ${\mathbf{r}}_1$ to $\mathbf{s}$. The law of cosines becomesDifferentiating and taking the wedge product with $du\wedge dr$ eliminates the terms involving $du$ and $dr$:Consequently the positive integration density transforms by $|\sin \theta d\theta |=vdv/(ur)$ at fixed $u,r$. The minus sign is accounted for by reversal of limits: $v$ decreases as $\theta$ increases. The radial variable retained is $r$, while the polar angle is replaced by $v$; equivalently one can start with the angle between the two position vectors and replace that angle by $r$. The wording of the printed hint conflates these two coordinate choices, but its volume form gives the stated transformation directly.

Integrating $\sin {\theta }_{1}d{\theta }_{1}d{\varphi }_1$ over the first orientation gives $4\pi$, and integrating $d\chi$ gives $2\pi$. The probability density function of $r$ is thereforewhere $D_r$ has $0\le u,v\le R$ and $|u-v|\le r\le u+v$. There are no such configurations for $r>2R$.

To evaluate this integral explicitly, the Jacobian determinant of $(u,v)=((x+y)/2,(x-y)/2)$ has absolute value $1/2$. Hence $uvdudv=(x^2-y^2)dxdy/8$ as a positive density. For $0\le r\le 2R$, set $a=\min (r,2R-r)$. The domain is $-a\le y\le a$, $r\le x\le 2R-|y|$, and evenness in $y$ givesAn antiderivative, zero at $y=0$, yieldsSubstituting $a=r$ for $r\le R$ and $a=2R-r$ for $r\ge R$ gives the same polynomial in both intervals:Thus the distance between two uniform points in a three-dimensional ball has the final densityIt is zero outside this interval. Its nonnegativity also follows from $f_R(r)=3r^2(4R+r)(2R-r{)}^2/(16R^6)$, and direct integration gives ${\int }_0^{2R}{f}_R(r)dr=1$. The derivation uses the full six-dimensional probability volume differential form, rather than treating the three scalar distances as independent.

The integral identity requires compactness of $M$, compact support of $X$, or suitable conditions eliminating flux at infinity. The printed question does not state such a hypothesis. We prove the Riemannian divergence theorem for a compact manifold, and more generally for a compactly supported smooth vector field on an oriented Riemannian manifold with boundary.

Write $j:\partial M\hookrightarrow M$ for the inclusion. Give $\partial M$ the outward-normal-first boundary orientation, and let $N$ be the outward unit normal vector. The Riemannian volume form of the induced Riemannian metric isIndeed, if $(e_1,\dots ,e_{n-1})$ is a positively oriented orthonormal basis tangent to the boundary, $(N,e_1,\dots ,e_{n-1})$ is a positively oriented orthonormal basis in $M$. Both sides therefore evaluate to one on the boundary basis.

At a boundary point decompose $X=g(X,N)N+X^{\top }$, where $X^{\top }$ is tangent to $\partial M$. Evaluating on $n-1$ tangent vectors, the term with $X^{\top }$ vanishes: all $n$ arguments of the alternating volume form then lie in the $(n-1)$-dimensional boundary tangent space. Thus the interior product of a differential form satisfiesThe general Stokes theorem for an oriented manifold with boundary states ${\int }_{M}d\eta ={\int }_{\partial M}{j}^{\ast }\eta$ for any smooth differential form $\eta$ of degree $n-1$ with compact support; compactness of $M$ makes the support condition automatic. Applying it to $\eta ={\iota }_X{\omega }_g$ and using the defining equation for the divergence of a Riemannian vector field proves

For a concrete failure without the extra hypothesis, take $M=[0,\infty )$ with Riemannian metric $d{x}^2$ and $X=(\arctan x){\partial }_x$. The divergence of a Riemannian vector field is $\mathrm{div}X=(1+x^2{)}^{-1}$, sobecause $X$ vanishes at the only boundary point. Both integrals are finite. The missing $\pi /2$ is the limiting flux at infinity, so mere integrability of the divergence cannot repair the unrestricted assertion.

Real line bundles over a paracompact space $X$ are classified byEuclidean space ${\mathbb{R}}^n$ is a contractible space, so its first cohomology vanishes and every real line bundle on it is trivial. Apply this to the defining bundle from part (c). In a global trivialization, $s$ is a smooth real function with $Y=s^{-1}(0)$ and $ds{|}_Y\ne 0$. Thus $ds$, or a metric-dual normal vector field, gives a global orientation of the normal line. Combining this with the standard orientation of ${\mathbb{R}}^n$ gives an orientation of $Y$ using the same normal-first convention as the outward-normal-first boundary orientation. Hence every properly embedded hypersurface in ${\mathbb{R}}^n$ is orientable.

Give $\partial X$ the outward-normal-first boundary orientation. The Generalized Stokes theorem states that, for every compactly supported $(n-1)$-form $\omega$,

Choose an oriented coordinate cover by charts into ${\mathbb{R}}^n$ or the half-space ${\mathbb{H}}^n=\{x^1\ge 0\}$, and choose a partition of unity $({\rho }_i)$ subordinate to it. Since the family is locally finite and $\omega$ has compact support, only finitely many ${\rho }_i\omega$ are nonzero. It is therefore legitimate to write both integrals as finite sums and prove the identity for a form supported in one chart.

In an interior chart the integral of an exact compactly supported top form is zero by the fundamental theorem of calculus. In a boundary chart writeIntegrating $d\omega =({\sum }_j{\partial }_j{a}_j)d{x}^1\wedge \cdots \wedge d{x}^n$ coordinate by coordinate kills every tangential derivative. The normal derivative leaves precisely the restriction to $x^1=0$, with the sign selected by the outward-normal-first convention. This is ${\int }_{\partial X}{F}^{\ast }\omega$, proving the theorem.

Let $({\theta }^1,\dots ,{\theta }^n)$ be the dual basis of the positively oriented orthonormal basis $(v_1,\dots ,v_n)$. By the definition of the Riemannian volume form,The interior product of a differential form with the outward unit normal isThe vectors $(v_2,\dots ,v_n)$ form a positive orthonormal frame of $T\partial X$ by the outward-normal-first boundary orientation. Consequently the pullback of the last display is the positive unit boundary volume form:

On an oriented Riemannian manifold with boundary, the divergence of a Riemannian vector field obeys the displayed integral identity for smooth compactly supported vector fields $X$, or for arbitrary smooth $X$ if the manifold is compact. Here $N$ is the outward unit normal vector and the boundary has the outward-normal-first boundary orientation. Its induced Riemannian volume form is $j^{\ast }{\iota }_N{\omega }_g$. Splitting $X$ into its normal and tangential components gives $j^{\ast }{\iota }_X{\omega }_g=g(X,N){\omega }_{\tilde{g}}$, since a top form vanishes on $n$ boundary-tangent arguments. Apply the Stokes theorem to $d{\iota }_X{\omega }_g=(\mathrm{div}X){\omega }_g$.

Without a support or decay condition, flux at infinity can invalidate the result even if the integrals converge. On $[0,\infty )$ with Riemannian metric $d{x}^2$, $X=(\arctan x){\partial }_x$ has boundary flux zero but integral of divergence $\pi /2$.