Integrating a differential form along a path

Question 1. The function $f(x,y) = xy$ works, since $\partial f/\partial x = y$ and $\partial f/\partial y = x$. Question 2. For a curve parametrized by $t\mapsto\bigl(x(t),y(t)\bigr)$, we have

$$\int_\gamma\omega_1 = \int \left[y(t)x'(t)+x(t)y'(t)\right]\,\mathrm{d}t.$$

Path $\gamma_1$ is the union of two segments. On the first, $x(t)=t$ and $y(t)=0$, with $t\in[0,1]$; hence $\mathrm{d}x=\mathrm{d}t$ and $\mathrm{d}y=0$. On the second, $x(t)=1$ and $y(t)=t$, again with $t\in[0,1]$; hence $\mathrm{d}x=0$ and $\mathrm{d}y=\mathrm{d}t$. Therefore,

$$\begin{aligned} I_1 = \int_{\gamma_1}\omega_1 &= \int_0^1\left(0\times 1+t\times 0\right)\,\mathrm{d}t + \int_0^1\left(t\times 0+1\times 1\right)\,\mathrm{d}t \\ &= 0+\int_0^1\mathrm{d}t = 1. \end{aligned}$$

For $\gamma_2$, the first segment is parametrized by $x(t)=0$, $y(t)=t$, and the second by $x(t)=t$, $y(t)=1$, with $t\in[0,1]$ in both cases. We therefore obtain

$$\begin{aligned} I_2 = \int_{\gamma_2}\omega_1 &= \int_0^1\left(t\times 0+0\times 1\right)\,\mathrm{d}t + \int_0^1\left(1\times 1+t\times 0\right)\,\mathrm{d}t \\ &= 0+\int_0^1\mathrm{d}t = 1. \end{aligned}$$

The two values agree. This could have been predicted without calculation: because $\omega_1=\mathrm{d}f$ with $f(x,y)=xy$, its integral depends only on the endpoints,

$$\int_\gamma\omega_1=f(M)-f(O)=f(1,1)-f(0,0)=1.$$

Question 3. Write $\omega_2=A(x,y)\,\mathrm{d}x+B(x,y)\,\mathrm{d}y$. Here $A(x,y)=y$ and $B(x,y)=0$. If the form were exact, Schwarz's criterion would give

$$\frac{\partial A}{\partial y}=\frac{\partial B}{\partial x}.$$

However, $\partial A/\partial y=1$, whereas $\partial B/\partial x=0$. The mixed derivatives differ, so $\omega_2$ is not exact. Its integral may therefore depend on the path followed between $O$ and $M$, as the following calculation confirms.

Question 4. Since $\omega_2=y\,\mathrm{d}x$, only a change in $x$ at a nonzero value of $y$ can contribute to the integral. Using the previous parametrizations, on $\gamma_1$ we find

$$\begin{aligned} J_1 =\int_{\gamma_1}\omega_2 &=\int_0^1 0\times 1\,\mathrm{d}t +\int_0^1 t\times 0\,\mathrm{d}t \\ &=0. \end{aligned}$$

On $\gamma_2$, the first segment satisfies $x(t)=0$, so $\mathrm{d}x=0$; on the second, $x(t)=t$, $y(t)=1$, and $\mathrm{d}x=\mathrm{d}t$. Thus

$$\begin{aligned} J_2 =\int_{\gamma_2}\omega_2 &=\int_0^1 t\times 0\,\mathrm{d}t +\int_0^1 1\times 1\,\mathrm{d}t \\ &=1. \end{aligned}$$

Finally, the diagonal segment $\gamma_3$ is explicitly parametrized by

$$x(t)=t,\qquad y(t)=t,\qquad t\in[0,1],$$

so $\mathrm{d}x=\mathrm{d}t$ and $\mathrm{d}y=\mathrm{d}t$. Hence

$$J_3 =\int_{\gamma_3}\omega_2 =\int_0^1 y(t)x'(t)\,\mathrm{d}t =\int_0^1 t\,\mathrm{d}t =\frac12.$$

The three paths have the same endpoints but yield three different values: $J_1=0$, $J_2=1$, and $J_3=1/2$. This is precisely the path dependence characteristic of a non-exact differential form.