Line Integrals and the Gradient Theorem: Examples and Counterexamples

By [L2], ${\gamma }^{\mathrm{\prime }}(t)=(-\sin t,\cos t)$ and $\parallel {\gamma }^{\mathrm{\prime }}(t){\parallel }_2=1$, while $f(\gamma (t))=\cos t$.

By [L1], [L2], and [L3], ${\int }_{\gamma }xds={\int }_{-\pi \mathrm{/}2}^{\pi \mathrm{/}2}\cos tdt=\sin (\pi \mathrm{/}2)-\sin (-\pi \mathrm{/}2)=2.$

Reversing the path leaves the value $2$ unchanged by [L4], confirming that the scalar $ds$ integral does not depend on orientation.

Differentiation gives ${\gamma }^{\mathrm{\prime }}(t)=q-p$, so the speed is the constant $\parallel q-p{\parallel }_2$.

Substitute step 1.1 and the formula for $\gamma (t)$ into the scalar clause of [L1]. Pulling out the constant speed gives the first displayed formula.

Substitute step 1.1 into the vector clause of [L1] to obtain the second displayed formula.

If $p=q$, then $q-p=0$ and $\parallel q-p{\parallel }_2=0$, so both formulas give zero. Thus the constant-segment case is included.

Coordinate differentiation gives the displayed gradient, and direct evaluation gives $\varphi (q)=4$ and $\varphi (p)=0$.

For the affine segment $\gamma (t)=(t,2t,t)$, one has $F(\gamma (t))=(4t^2,2t^2,4t^2)$ and $q-p=(1,2,1)$. Thus [L2] gives the integrand $12t^2$.

Apply [L1] and step 1.1 to any path from $p$ to $q$. Its integral is $4-0=4$.

Since $12t^2=(4t^3{)}^{\mathrm{\prime }}$, [L3] gives ${\int }_0^{1}12t^{2}dt=4$, agreeing with step 2.1.

Direct differentiation gives ${\mathrm{\partial }}_y{F}_0=2x+2y={\mathrm{\partial }}_x{F}_1,$ so $F$ is closed by [L1].

Evaluating the two polynomial integrals using [L4] gives $\varphi (x,y)=b(x^2-a^2)+b^2(x-a)+x^2(y-b)+x(y^2-b^2)=x^{2}y+x{y}^2-a^{2}b-a{b}^2\mathrm{.}$

Differentiating step 1.2 yields ${\mathrm{\partial }}_x\varphi =2xy+y^2=F_0$ and ${\mathrm{\partial }}_y\varphi =x^2+2xy=F_1$. Thus [L1] makes $\varphi$ a potential, and substitution gives $\varphi (a,b)=0$.

An open rectangle is convex and hence star-shaped with respect to $(a,b)$, so [L2] also supplies a radial potential normalized to zero there. By [L3] and the common normalization, that radial potential agrees with $\varphi$.

Both paths run from $(0,0)$ to $(1,0)$. Along $\alpha$, the first component $y$ is zero, so [L1] gives ${\int }_{\alpha }F\cdot d\mathbf{r}=0$.

Along $\beta$, one has $F(\beta (t))=(t(1-t),0)$ and ${\beta }^{\mathrm{\prime }}(t)=(1,1-2t)$. Hence [L1] and [L4] give ${\int }_{\beta }F\cdot d\mathbf{r}={\int }_0^1(t-t^2)dt=\frac{1}{6}\mathrm{.}$

The two values differ, so the field is not path-independent.

The concatenation $\alpha \ast {\beta }^{-}$ is closed, and [L2] gives its integral as $0-1\mathrm{/}6=-1\mathrm{/}6$. This is the corresponding nonzero-loop failure in [L3].

The rational formulas defining $F$ are $C^1$ on $U$. Direct differentiation gives ${\mathrm{\partial }}_y{F}_0=\frac{y^2-x^2}{(x^2+y^2{)}^2}={\mathrm{\partial }}_x{F}_1,$ so $F$ is closed by [L1].

The punctured plane is piecewise-$C^1$ path-connected: choose a positive radius at least as large as the radii of two given points, join each point outward along its own ray to that circle, and join the resulting points by a circular arc. None of these pieces meets the origin.

On the counterclockwise unit circle $\gamma (t)=(\cos t,\sin t)$, $0\le t\le 2\pi$, [L5] gives $F(\gamma (t))=(-\sin t,\cos t)={\gamma }^{\mathrm{\prime }}(t)\mathrm{.}$

By [L4], [L5], and [L6], ${\int }_{\gamma }F\cdot d\mathbf{r}={\int }_0^{2\pi }({\sin }^{2}t+{\cos }^{2}t)dt=2\pi \mathrm{\ne }0.$

If $F$ were exact, [L1] would give a potential and [L2] would make the closed-circle integral zero, contradicting step 2.1. Thus $F$ is closed but not exact. By [L3] and step 1.2, it is also neither conservative nor path-independent.

The domain is not star-shaped: for any proposed centre $a\mathrm{\ne }0$, the segment from $a$ to $-a$ passes through the omitted origin.

Direct differentiation gives ${\mathrm{\partial }}_y{F}_0=(y^2-x^2)\mathrm{/}(x^2+y^2{)}^2={\mathrm{\partial }}_x{F}_1$, so [L1] makes $F$ closed and $C^1$ on the open set $U$. Radial segments at positive radius followed by a circular arc give a piecewise-$C^1$ path in $U$ between any two of its points.

For $\gamma (t)=(\cos t,\sin t)$ on $[0,2\pi ]$, [L3] gives $F(\gamma (t))={\gamma }^{\mathrm{\prime }}(t)=(-\sin t,\cos t)$. Hence [L3] and [L4] give ${\int }_{\gamma }F\cdot d\mathbf{r}={\int }_0^{2\pi }1dt=2\pi \mathrm{\ne }0.$

If $U=A\cup B$ were a separation into disjoint nonempty relatively open sets, choose $u\in A$ and $v\in B$ and let $\eta :[0,1]\to U$ be the path from step 1.1. The function equal to $0$ when $\eta (t)\in A$ and to $1$ when $\eta (t)\in B$ is locally constant, hence continuous, and has endpoint values $0$ and $1$. By [L5] it would take the value $1\mathrm{/}2$, which is impossible. Thus $U$ is connected.

If $F$ were exact, [L1] and [L2] would make the integral in step 1.2 zero. Thus it is not exact, and the statement is false despite steps 1.1 and 2.1. The valid correction in [L6] replaces connectedness by the stronger star-shaped hypothesis.

Since ${\gamma }^{\mathrm{\prime }}(t)=(1,0)$, [L1] and [L3] give ${\int }_{\gamma }F\cdot d\mathbf{r}={\int }_0^{1}1dt=1$.

The reversal is orientation-reversing, so [L2] gives ${\int }_{{\gamma }^{-}}F\cdot d\mathbf{r}=-1$.

Since $-1\mathrm{\ne }1$, the statement is false. The scalar analogue is true because [L2] says that $ds$ removes the orientation sign.

By [L2], $F(\gamma (t))={\gamma }^{\mathrm{\prime }}(t)=(-\sin t,\cos t)$, so [L1], [L2], and [L3] give ${\int }_{\gamma }F\cdot d\mathbf{r}={\int }_0^{2\pi }1dt=2\pi$.

By [L2], ${\beta }^{\mathrm{\prime }}(t)=2(-\sin (2t),\cos (2t)),F(\beta (t))=(-\sin (2t),\cos (2t))\mathrm{.}$

Hence [L1], [L2], and [L3] give ${\int }_{\beta }F\cdot d\mathbf{r}={\int }_0^{2\pi }2({\sin }^2(2t)+{\cos }^2(2t))dt=4\pi \mathrm{.}$

The two paths have the same counterclockwise unit-circle image, but $t\mapsto 2t$ on $[0,2\pi ]$ is not a bijection onto $[0,2\pi ]$; it covers the circle twice. Thus [L4] does not assert equality here.