Evaluate the line integral separately on the three oriented line segments. Their parametrizations and contributions areThusThe different value in i demonstrates path dependence. The vector field in this part is different from the conservative vector field in part (a).

A parametrization of the straight line segment is $\mathbf{r}(t)=(t,t,t)$, $0\le t\le 1$. Then $\mathbf{F}(\mathbf{r}(t))=(t,t^2,t)$ and ${\mathbf{r}}^{\prime }(t)=(1,1,1)$. By the line integral formula,

The inverse change of variables is $x=v/(1+u)$, $y=uv/(1+u)$. Thus the triangular region becomesThe upper limit depends on $u$: the new region is not a rectangle. The absolute inverse Jacobian determinant is $x^2/(x+y)$, so the change of variables formula cancels the rational factor in the integrand:Although the original integrand is undefined at the vertex $x=y=0$, its nonnegative improper integral exists. One can apply the change of variables formula on $x\ge \epsilon$ and let $\epsilon \downarrow 0$ by the monotone convergence theorem; the transformed integrand has the finite integral just computed.

The Jacobian determinant isThe change of variables is defined only for $x\ne 0$. On the interior of the integration region, $x>0$ and $y>0$, so this Jacobian determinant is negative and nonzero.

For the stated inward heating, take $Q\ge 0$ and put $\tau ={\pi }^{2}kt/(16{\ell }^2)$. Since $|\cos ({\mu }_{n}x)|\le 1$, the transient from ii satisfiesEquality holds at $x=0$, where every cosine is one and all coefficients have the same sign. Thus the maximum is attained at the heated end, and its exact value and late-time form areThis is dominance of the slowest-decaying eigenfunction when $kt\gg {\ell }^2$. More explicitly, the omitted relative correction is nonnegative and bounded by $({\pi }^2/8-1)e^{-8\tau }$, using ${\sum }_{n\ge 0}(2n+1{)}^{-2}={\pi }^2/8$. For a signed imposed heat flux, the maximum-deviation formula uses $|Q|$.

Write $T=T_s+T_u$. The transient satisfies the heat equation and homogeneous mixed boundary conditionsThe left condition is a Neumann boundary condition and the right one a Dirichlet boundary condition. Separation of variables, $T_u=X(x)G(t)$, givesThe condition $X^{\prime }(0)=0$ removes the sine term. The condition $X(2\ell )=0$ then selectsThere is no nonzero mode with $\mu =0$, because a linear $X$ satisfying both conditions is zero. There is no growing mode either: integration by parts gives ${\int }_0^{2\ell }(X^{\prime }{)}^{2}dx={\mu }^2{\int }_0^{2\ell }{X}^{2}dx$, so every nonzero mode has ${\mu }^2>0$.

These Sturm-Liouville eigenfunctions have orthogonality and squared normExpand the initial transient in this half-integer Fourier cosine series. Its coefficients areThe integration formula follows by integration by parts; the cosine at the endpoint is zero. Thus the half-integer cosine expansion of a linear profile givesAt $t=0$ the Fourier series reproduces $-T_s$ everywhere, including the endpoints; its $n^{-2}$ coefficient decay gives uniform convergence. For every $t>0$ the exponential factors allow termwise differentiation and verify the heat equation and both boundary conditions. The sudden switch-on of nonzero heat flux is incompatible with the spatial derivative of the constant initial temperature at the corner $(0,0)$. Accordingly, the positive-time boundary derivative must not be inferred by differentiating the initial Fourier series term by term at that corner.

Inward heat flux at the left end points in the positive $x$ direction, so the boundary conditions areFor uniform thermal conductivity, the steady heat equation gives $T_s^{\prime \prime }=0$. The heat flux is constant and equal to $Q$, so $T_s^{\prime }=-Q/k$. Imposing the right-end temperature gives

The truncated Fourier cosine series gives the better approximation over the whole closed interval. Its Fourier coefficients obey $|a_n|\le 4/(\pi n^2)$, so absolute convergence and the tail boundgive uniform convergence. The even periodic extension is continuous; its slope changes only cause corners.

By contrast $b_n=(-1{)}^{n+1}/n+O(n^{-2})$. The odd periodic extension has a jump at $\pi$, giving slower coefficient decay and the Gibbs phenomenon near that endpoint. Every truncated Fourier sine series satisfies $S_N(\pi )=0$, soIt therefore cannot converge uniformly to $f$ on $[0,\pi ]$, although it converges at every point of $[0,\pi )$. This comparison concerns approximation across the interval; it does not assert that the cosine truncation has smaller error at every individual interior point for every $N$.

The Fourier cosine series converges to the continuous even $2\pi$-periodic extension. If $r\in [-\pi ,\pi ]$ represents $x$ modulo $2\pi$, its value isIt has value zero at $x=2j\pi$, rises linearly with distance from these points until height $\pi /2$, and then stays flat until the next descending ramp. Thus the sketch on $[-3\pi ,3\pi ]$ has minima at $-2\pi ,0,2\pi$, and plateaux centred at the odd multiples of $\pi$. Its large partial sums round the corners and closely follow the whole graph.

The Fourier sine series converges, away from the odd multiples of $\pi$, to the odd $2\pi$-periodic extensionAt every odd multiple of $\pi$ its value is zero, the midpoint of the jump from $\pi /2$ to $-\pi /2$. At even multiples of $\pi$ it passes continuously through zero. Thus on every interval $((2j-1)\pi ,(2j+1)\pi )$ it has a negative plateau, a rising linear segment through $2j\pi$, and a positive plateau. The jump size is $\pi$.

The blue curves are the partial sums through $n=60$. The open endpoint circles and filled zero dots in the lower panel distinguish the one-sided limits from the value of the Fourier sine series at each jump. Near these jumps the partial sums exhibit the Gibbs phenomenon: the oscillations become concentrated in narrower regions as the number of terms grows, while their limiting overshoot remains about $0.08949$ times the jump, here about $0.281$.

Put $a=\pi /2$. The constant Fourier coefficient isFor $n\ge 1$, integration by parts givesThe $n^{-1}$ terms cancel. ThusThis Fourier cosine series has $a_n=0$ when $n$ is divisible by four, $a_n=-4/(\pi n^2)$ when $n\equiv 2(\mathrm{mod}4)$, and $a_n=-2/(\pi n^2)$ when $n$ is odd.

Similarly,ThereforeThe Fourier sine series also equals $f(0)=0$ at zero, but has value zero at $\pi$, where $f(\pi )=\pi /2$.

Using the usual piecewise smooth interpretation of the derivative hypothesis, Dirichlet's theorem on Fourier series gives the average of the two one-sided limits of the appropriate periodic extension. The even extension is continuous at $0$ and at the identified endpoints $\pm \pi$, so the Fourier cosine series converges toFor the odd extension, the one-sided limits at zero are $-f(0)$ and $f(0)$, while at $\pi$ the periodic one-sided limits are $f(\pi )$ and $-f(\pi )$. Hence the Fourier sine series converges toAt an endpoint this equals $f$ precisely when $f$ is zero there. One can also see the zero endpoint values directly: every partial sum consists of sines that vanish at both endpoints.

For the Fourier cosine series, first take the even extension $F_e(x)=f(|x|)$ on $[-\pi ,\pi ]$, and then its $2\pi$-periodic extension. Evenness eliminates the sine terms. With the constant term written as $a_0/2$, the Fourier coefficients and series areFor the Fourier sine series, take the odd extension $F_o(x)=f(x)$ for $0<x<\pi$ and $F_o(x)=-f(-x)$ for $-\pi <x<0$, and extend with period $2\pi$. Values assigned at isolated endpoints do not change the Fourier coefficients. Oddness eliminates the constant and cosine terms, leavingBoth formulas follow by orthogonality on $[-\pi ,\pi ]$ and the parity of the extensions.

Parametrize the straight segment by $\mathbf{r}(t)=(t,t,t)$, $0\le t\le 1$. Then the line integral isThe three different answers are possible because this vector field is not a conservative vector field; for example ${\partial }_y{F}_x=6$ whereas ${\partial }_x{F}_y=0$.

Compute the line integral on each straight segment. On the first, $y=z=0$ and only $dx$ contributes; on the second, $x=1,z=0$ and only $dy$ contributes; on the third, $x=y=1$ and only $dz$ contributes. ThereforeAlthough $F_x=9$ on the second segment, that term does no work there because $dx=0$.

For the parametric curve $\mathbf{r}(t)=(t,t^2,t^3)$, $0\le t\le 1$, the vector field and the tangent vector areConsequently the line integral is

An orientation-preserving reparametrization has the form $\tilde{\gamma }(s)=\gamma (h(s))$, where $h:[\alpha ,\beta ]\to [a,b]$ is an increasing, piecewise smooth change of parameter with $h(\alpha )=a$, $h(\beta )=b$. For regular parametrizations one may take $h$ and its inverse continuously differentiable on each piece. The chain rule gives ${\tilde{\gamma }}^{\prime }(s)={\gamma }^{\prime }(h(s))h^{\prime }(s)$. The change of variables $t=h(s)$ therefore yieldsThis proves reparametrization invariance of a vector line integral. If $h$ is decreasing, the endpoints in the final integral are reversed and the sign changes. Thus the same direction and the same multiplicity of traversal are essential.

For a continuous vector field along an oriented piecewise continuously differentiable parametric curve $\gamma :[a,b]\to {\mathbb{R}}^3$, define its line integral byadding the integrals over the smooth pieces when necessary. The value refers to the specified traversal: going around a curve twice counts twice.

The displayed vector field is defined on the punctured plane $U={\mathbb{R}}^2\setminus \{(0,0)\}$; it is undefined at the origin. Take the counterclockwise unit circle $\mathbf{r}(t)=(\cos t,\sin t)$, $0\le t\le 2\pi$. Along it,By the fundamental theorem for line integrals, it has no single-valued potential of a conservative vector field on $U$. In particular it cannot be the restriction of the gradient of a smooth function on all of ${\mathbb{R}}^2$.

For $(x,y)\ne (0,0)$, direct differentiation givesThere is no contradiction with Green's theorem: $\mathbf{F}$ is not defined, let alone continuously differentiable, throughout the disk enclosed by $C$. On an annulus the theorem does apply; the counterclockwise outer integral is $2\pi$, and the clockwise inner integral is $-2\pi$, so their sum is zero. The hole prevents this curl-free vector field from being a conservative vector field on its entire domain.

For an oriented piecewise smooth parametric curve $\mathbf{r}:[a,b]\to {\mathbb{R}}^3$ and a continuous vector field $\mathbf{F}$ along it, the vector-field line integral isFor a piecewise smooth curve, sum these integrals over its smooth pieces. The dot product selects the component of the vector field in the direction of motion.

If $\mathbf{F}=\nabla \varphi$ for a smooth function $\varphi$, the multivariable chain rule givesThe fundamental theorem of calculus, with cancellation at intermediate endpoints, therefore proves the fundamental theorem for line integrals:On a closed curve $\mathbf{r}(a)=\mathbf{r}(b)$, so $\boxed{{\oint }_C\nabla \varphi \cdot d\mathbf{r}=0}$.

In polar coordinates, put $x=r\cos \theta$ and $y=r\sin \theta$. Differentiating along the curve givesThe cardioid has $r\ge 0$ and is traversed once counterclockwise as $\theta$ increases from $0$ to $2\pi$. Its cusp at $\theta =\pi$ causes no difficulty: split the line integral there, or exclude a small arc and take its limit. Thus signed area as a boundary integral gives