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A differential form is a smooth field of alternating multilinear maps that can be integrated over an appropriately oriented manifold of matching dimension. A one-form measures a directed displacement; a two-form measures an oriented area element. The point is to express what is being accumulated along a curve or over a surface without tying the answer to one set of coordinates. This makes line integrals, surface integrals, and the integral theorems of vector calculus parts of one construction.

For a concrete one-form, take $\alpha =xdy$ in the plane. Travel from $(0,0)$ to $(1,1)$ along the bottom and then the right edge of the unit square. The horizontal leg contributes zero because $dy=0$; the vertical leg has $x=1$ and contributes ${\int }_0^{1}dy=1$. Taking the left and then the top edge instead gives zero: the vertical leg has $x=0$, and the horizontal leg again has $dy=0$. The two paths have the same endpoints but different integrals.

Now replace $\alpha$ with the exact differential $\beta =d(xy)=ydx+xdy$. Both paths give $1$, because integrating an exact differential along a curve gives the endpoint difference of its potential. Here that difference is $xy{|}_{(1,1)}-xy{|}_{(0,0)}=1$. This is the fundamental theorem of calculus applied to a function along a parametric curve, and is the basic reason conservative vector fields have path-independent work.

The exterior derivative records the local obstruction in the first example: $d(xdy)=dx\wedge dy$. The Generalized Stokes theorem turns the difference between the two path integrals into the integral of this two-form over the square, whose oriented area is $1$. Interior edges cancel when adjoining pieces are added; only the outer boundary survives. That cancellation connects a local calculation with a global measurement.

More precisely, a degree-$k$ differential form assigns to each point $p$ a multilinear scalar-valued function of $k$ tangent vectors at $p$. Interchanging two arguments changes its sign. A one-form is therefore a covector field, while a two-form vanishes on a pair of parallel vectors. Degree zero simply means a smooth function. The space ${\Omega }^k(M)$ is a vector space; the wedge product of differential forms combines degrees, and the exterior derivative raises degree by one.

The usefulness goes beyond bookkeeping. Pullbacks of a differential form supply the right transformation law automatically under a coordinate change. The Generalized Stokes theorem unifies the fundamental theorem of calculus, Green theorem, Stokes theorem, and divergence theorem. Meanwhile, the distinction between closed differential forms and exact differential forms detects holes through de Rham cohomology. Reyer Sjamaar develops these connections in sections 5.4 and 9.3 of his university notes. ^[ref].

A differential one-form on a specified domain is inexact if it is not the differential of a single-valued scalar potential there. For $Pdx+Qdy$, unequal cross derivatives $P_y\ne Q_x$ prove inexactness locally. Equality is sufficient on a simply connected plane domain under the usual smoothness hypotheses, but holes can obstruct a global potential. The form $42ydx+xdy$ has cross derivatives $42$ and $1$ and counterclockwise unit-square integral $-41$, so it is inexact. A conservative vector field has an exact differential as its work form.

For a simple compressible system at fixed particle number, assume the internal energy has continuous second partial derivatives and $T$ is a locally invertible function of entropy at fixed volume. Its exact differential $dU=TdS-pdV$ gives the Maxwell relation $T_V{|}_S=-p_S{|}_V$. Meanwhile the constant-volume heat capacity is $C_V=T{S}_T{|}_V$. The chain rule therefore givesFor one mole of an ideal gas with constant molar $C_V>0$, this becomes $T_V{|}_S=-RT/(C_{V}V)$, whose integral curves satisfy $T{V}^{R/C_V}=\text{constant}$. It expresses cooling on reversible adiabatic expansion in terms of measurable thermal and mechanical derivatives.

An exact differential $Mdx+Ndy$ is $d\Phi$ for a scalar potential function $\Phi$, so $M=\partial \Phi /\partial x$ and $N=\partial \Phi /\partial y$. Equality $\partial M/\partial y=\partial N/\partial x$ is necessary, and is sufficient on a simply connected space when these partial derivatives are continuous.

Here, after multiplying the ordinary differential equation by $dx$,On either half-plane $y>0$ or $y<0$, integrate $M$ with respect to $x$:Comparison of $\partial \Phi /\partial y$ with $N$ gives $h^{\prime }(y)=6y$, hence $h(y)=3y^2$ up to a constant. The solution is therefore the family of level setsLocally these define differentiable graphs $y(x)$ wherever $N\ne 0$, by the implicit function theorem. When $N=0$, $M\ne 0$ on this domain, so a finite derivative cannot satisfy $M+N{y}^{\prime }=0$ there; such points are vertical tangents of the implicit curves, rather than additional classical graph solutions.

For the given polynomial vector field, direct differentiation givesAn exact differential is a differential one-form equal to $d\varphi ={\varphi }_{x}dx+{\varphi }_{y}dy+{\varphi }_{z}dz$ for a single-valued scalar function. This field is smooth on the simply connected domain ${\mathbb{R}}^3$, so its vanishing curl is sufficient for exactness. More concretely,The explicit potential proves exactness globally, rather than relying on curl-free behavior without checking the domain.

The field is continuously differentiable on all of ${\mathbb{R}}^3$, a simply connected domain. Its vanishing curl therefore implies a global potential of a conservative vector field, so $\mathbf{F}\cdot d\mathbf{r}=d\varphi$ is an exact differential. This global domain condition matters: curl-free fields on domains with holes need not have single-valued potentials.

Integrating ${\varphi }_x=2y{z}^2$ gives $\varphi =2xy{z}^2+g(y,z)$. Matching ${\varphi }_y=2x{z}^2$ forces $g_y=0$, so $g=h(z)$; matching ${\varphi }_z=4xyz$ then forces $h^{\prime }=0$. ConsequentlyThe additive constant does not affect the field or any endpoint difference.

A one-form $A(x,y)dx+B(x,y)dy$ is an exact differential if a single-valued differentiable potential of a conservative vector field $f$ satisfies $A=f_x$ and $B=f_y$ on the domain. Then it equals $df$, and its integral along a path is the endpoint difference of $f$.

Integrate $f_x=10x^4-2x{y}^3$ with respect to $x$:Its $y$ derivative is $-3x^2{y}^2+g^{\prime }(y)$. Matching the required second component forces $g^{\prime }(y)=0$, so the additive constant is arbitrary andThe fundamental theorem for line integrals now givesindependently of the convenient segmented path used before. Merely checking equality of mixed partial derivatives on an unspecified domain would not by itself construct this globally defined potential.

By Stokes theorem, it suffices to calculate the boundary line integral. On the $xz$ plane, $y=0$ and $dy=0$, soThis exact differential has zero integral around a closed curve. ThereforeThe argument applies to any admissible spanning surface, including a curved one; the surface itself need not lie in the $xz$ plane.

For an exact differential $Mdx+Ndy$, equality of the mixed partial derivatives requires $M_y=N_x$. Here these are $\sin x+b\cos y$ and $\sin x+3\cos y$, so $\boxed{b=3}$. Integrating $M$ with respect to $x$ gives $f=-y\cos x+3x\sin y+g(y)$. Comparison with $N$ gives $g^{\prime }=y^2$, and the specified value fixes the integration constant. ThereforeDifferentiating recovers both coefficients, and $f(0,2)=-2+8/3-2/3=0$. The explicit potential function also proves exactness globally, without relying only on a local compatibility test.

A conservative vector field can be written as $F=\nabla \Phi$ for a single-valued scalar potential $\Phi$. Its line integral between two points depends only on the endpoints, and its integral around every closed curve is zero. A necessary local test for a smooth planar field $F=(P,Q)$ is $P_y=Q_x$. On a simply connected domain such as the whole plane, this is also sufficient.

HereAn explicit scalar potential is even more decisive:Indeed ${\Phi }_x=2x{e}^{xy}+x^{2}y{e}^{xy}$ and ${\Phi }_y=x^3{e}^{xy}+2y$. Thus the field is conservative on ${\mathbb{R}}^2$.

An exact differential is a differential one-form equal to $d\Phi$ on its domain; an inexact differential is not the differential of a single-valued potential there. For example, $d(xy)=ydx+xdy$ is exact. For the supplied form $\eta =42ydx+xdy$, the cross derivatives areThey disagree, so $\eta$ is not even closed and hence cannot be exact. Equivalently, its counterclockwise line integral around the unit square is $1-42=-41\ne 0$. ThereforeThe connection is $F\cdot dr=d\Phi$: a vector field is conservative precisely when its work differential one-form is exact. Locally vanishing curl alone need not guarantee global exactness on a domain with holes.

An exact differential is an expression $Pdx+Qdy$ that equals $df=f_{x}dx+f_{y}dy$ for a single-valued differentiable function $f$ on the domain. Its line integral is an endpoint difference by the fundamental theorem for line integrals.

Integrating $f_x=P$ with respect to $x$ gives $f=2x^5-x^2{y}^3+g(y)$. Matching $f_y=-3x^2{y}^2+g^{\prime }(y)$ to $Q$ requires $g^{\prime }(y)=0$. HenceThe arbitrary additive constant cancels. This constructs the vector-field potential globally and provides a second proof of path independence.

For an exact differential $Pdx+Qdy$, equality of mixed partial derivatives requires $P_y=Q_x$. The matching pair is $P=f_1$, $Q=f_3$, sinceThe other pairings do not give the same partial derivatives on an open region. Integrating $f_1$ with respect to $x$ gives $\ln x\sin y+e^x{\cos }^{2}y+C(y)$. Its $y$ derivative agrees with $f_3$ precisely when $C^{\prime }(y)=0$. HenceThis also proves existence, rather than relying only on the necessary mixed-partial derivative test.

Let $F^{\prime }=f$ and $G^{\prime }=g$ on the relevant intervals. The chain rule givesThus the one-form is explicitly an exact differential. Fixing its potential to vanish at $(1,1)$ determines it uniquely on the connected domain:Continuity of $f,g$ on the integration ranges suffices. For arbitrary smooth data defined throughout $x>0$, that half-plane also supplies the simply connected domain in the previous argument.

For the specialization $f(s)=s^{-2}$, the potential based at $(1,1)$ is defined in the positive quadrant; it cannot be continued through $xy=0$ using this formula. Taking $a,b>0$ keeps both endpoints in that component. Their ratios $y/x$ are both $b/a$, so the $G$ term cancels andNo condition on $g$ beyond existence on the traversed ratios enters this difference.

Consider the globally smooth potential $\Phi (x,y,z)=x{y}^2\sin (xz)$. Direct differentiation givesThus the differential one-form is the exact differential $d\Phi$, proving exactness rather than only checking a necessary condition on its partial derivatives. Along any piecewise differentiable path, the chain rule gives $udx+vdy+wdz=d\Phi$. The fundamental theorem of calculus therefore makes the line integral depend only on its endpoints:

An integrating factor for a differential one-form is a nonzero function $\mu (x,y)$ for which $\mu (dy+fdx)$ is an exact differential, say $dF$. Then solutions lie on level sets $F=\text{constant}$. Locally, the compatibility condition is ${\partial }_x\mu ={\partial }_y(\mu f)$.

Here multiplication by $2y{e}^x$, which is nonzero for $y>0$, producesThe initial condition makes this first integral equal to $a^2$. Taking the positive branch yields the solutionon the connected interval containing zero where the radicand is positive. This domain restriction matters because the original differential equation is singular at $y=0$.

Completing the square gives $2x+x^2=(x+1{)}^2-1\ge -1$. As $y>0$, the differential equation consequently impliesAt $x=0$, $y^{\prime }=-a/2<0$. Thus, moving a little to the left from $y(0)=a\ge 1$, the curve rises above $a$. Throughout $y>1$ every segment of the direction field points downwards as $x$ increases, by the inequality above. Tracing backwards therefore makes $y$ rise further, so it cannot return to $y=1$. This graphical barrier argument proves $y>a\ge 1$ and $y^{\prime }<0$ for every $x<0$ on the solution. The explicit first integral also shows this branch extends throughout $x<0$: writing $t=-x>0$, the minimum of $e^t/t^2$ is $e^2/4>1$, so $a^2{e}^t-t^2>0$ for $a\ge 1$. Including the initial point, $y^{\prime }<0$ for all $x\le 0$.

For $a=1$ the curve is $y=\sqrt{e^{-x}-x^2}$ on $(-\infty ,b)$, where the strictly increasing function $x^2{e}^x$ has the unique positive root $x^2{e}^x=1$ at $b\simeq 0.703467$. The curve decreases through $(0,1)$ with slope $-1/2$ and reaches $(b,0)$ only as a limiting endpoint. Its derivative isAs $x\to -\infty$, $y\sim e^{-x/2}$ and $y^{\prime }\sim -\frac{1}{2}{e}^{-x/2}\to -\infty$. As $x\uparrow b$, the numerator tends to $-b^2-2b<0$ and the denominator to zero from above, so again $y^{\prime }\to -\infty$. More precisely, $y\sim \sqrt{b(b+2)(b-x)}$ at the endpoint. Both ends have the requested unbounded negative slope.

For an exact differential $Md\theta +Nd\varphi =dF$, equality of mixed partial derivatives requires $M_{\varphi }=N_{\theta }$. Here the two partial derivatives areThey are not identically equal, so the differential form is not a closed differential form and hence is not an exact differential. Equality along special lines such as $\theta =0$ does not supply a potential on a two-dimensional open neighborhood, because every such neighborhood contains points where the two partial derivatives differ.

To construct a potential for the exact differential, first integrate its $dx$ coefficient with respect to $x$:Matching the $dy$ coefficient gives $g_y=-2z^2$, hence $g=-2y{z}^2+h(z)$. The remaining partial derivative is thenMatching the $dz$ coefficient forces $h^{\prime }=0$. Therefore a potential, valid throughout ${\mathbb{R}}^3$, isDifferentiating this expression recovers all three coefficients, proving that the differential form really is an exact differential.

An exact differential is a one-form equal to $d\Phi$ for a single-valued scalar function $\Phi$ on the domain. In Euclidean coordinates this means its vector coefficients form $\nabla \Phi$. Integrating the first component of $F$ with respect to $x$, then matching the other components, givesDifferentiating checks all three components. Since the domain ${\mathbb{R}}^3$ is connected, two such potentials differ only by a constant, so this is the most general potential.

At both specified endpoints the nonconstant terms vanish. The fundamental theorem for line integrals gives zero for the line integral along every path joining the two points.

For the planar field, choose a continuous lift $\theta (t)$ of the polar angle along a curve avoiding the origin. Differentiating the angle givesOn the closed curve the angle increases by $2\pi$ times its winding number. ThusThe angle differential is locally exact but not globally exact on the punctured plane: the angle lift need not return to its initial value. This is the nonexact angular one-form, whose nonzero closed-path integral obstructs a global potential.

When $\alpha =2\beta$, all three values reduce to $-\beta /3$. To prove path independence for every piecewise smooth path, construct a global potential rather than infer it from agreement on only three paths. Integrating the coefficient of $dx$ and then matching the coefficient of $dy$ givesIndeed ${\Phi }_x=2\beta (x^2+xy)=\alpha (x^2+xy)$ and ${\Phi }_y=\beta (x^2+y^2)$. The integrand is the exact differential $d\Phi$ throughout ${\mathbb{R}}^2$. The fundamental theorem for line integrals therefore gives, for every such path from $A$ to $B$,

A differential one-form $Pdx+Qdy$ is an exact differential if there is a differentiable function $F$ with $P=F_x$ and $Q=F_y$, so the form equals $dF$. For continuously differentiable $P,Q$, equality of mixed partial derivatives gives the necessary conditionA local differential condition need not by itself give a global potential on a domain with holes; no such issue arises for the explicit primitives here.

For the first form, $P_y=2y=Q_x$ and direct differentiation givesFor the second, $P_y=2y$ but $Q_x=y^2$. These are not equal on any nonempty open subset of the plane, so that form is not an exact differential. Nevertheless,Thus one suitable pair is $f(x,y)=x{e}^y$, $g(y)=y^2{e}^{-y}$. This factorization is smooth even at $y=0$. On regions where $y\ne 0$, its reciprocal factor $e^y/y^2$ is an integrating factor which turns the original form into $df$.