Line integrals in differential form

Question

I'm a bit confused as to the format of line integrals in differential form (i.e. the form in which Green's theorem is often presented). For example: $$ \oint\limits_\mathcal{C} \left( y^2 \mathrm{d}x + x^2 \mathrm{d}y \right)$$ where $\mathcal{C}$ is an anti-clockwise square joining $(0, 0),$ $(1, 0),$ $(1, 1)$ and $(0, 1)$.

To evaluate this without invoking Green's theorem, I rewrite this as $$ \oint\limits_\mathcal{C} y^2 \mathrm{d}x + \oint\limits_\mathcal{C}x^2 \mathrm{d}y $$ but the form of this makes me very uneasy. What does it mean to integrate along a two-dimensional path $\mathcal{C}$ with respect to a one-dimensional axis?

Is it correct to interpret this as following some arbitrary path, along which only the infinitesimal displacement in $x$ is considered for an infinitesimal progression of the path?

Is it therefore perfectly legal to only consider the differential change in a particular axis across a multidimensional path (which could be one hundred dimensions!)?

Answer 1

In general, we parameterize a smooth curve $C$ with $\vec r(t)=\hat xx(t)+\hat yy(t)$, $t\in[0,1]$, such that

$$\int_C\,\left(f(x,y)dx+g(x,y)dy\right)=\int_0^1 \left(f(x(t),y(t))\,\frac{dx(t)}{dt}+g(x(t),y(t))\,\frac{dy(t)}{dt}\right)\,dt$$

In the example at hand, we parameterize each line segment of $C$ separately. To that end, we have

$$\begin{align} &x=t, y=0,\,\,\text{for the segment from}\,\,(0,0)\,\,\text{to}\,\,(1,0)\\\\ &x=1, y=t\,\,\text{for the segment from}\,\,(1,0)\,\,\text{to}\,\,(1,1)\\\\ &x=-t, y=1\,\,\text{for the segment from}\,\,(1,1)\,\,\text{to}\,\,(0,1)\\\\ &x=0, y=-t\,\,\text{for the segment from}\,\,(0,1)\,\,\text{to}\,\,(0,0) \end{align}$$

The rest is just carrying out the four line integrals separately.

For the integral over the first segment, we have $x=t$, $\frac{dx}{dt}=1$, $y=0$, and $\frac{dy}{dt}=0$. Thus,

$$\int_0^1 ((0^2)(1)+(t)^2(0))dt=0$$

For the integral over the second segment, we have $x=1$, $\frac{dx}{dt}=0$, $y=t$, and $\frac{dy}{dt}=1$. The integral is $1$.

For the integral over the third segment, we have $x=-t$, $\frac{dx}{dt}=-1$, $y=1$, and $\frac{dy}{dt}=0$. The integral is $-1$.

For the integral over the forth segment, we have $x=0$, $\frac{dx}{dt}=0$, $y=-t$, and $\frac{dy}{dt}=-1$. The integral is $0$.

Finally, adding the contributions from the four path integrals reveals

$$\bbox[5px,border:2px solid #C0A000]{\oint_C (y^2dx+x^2dy)=0}$$

Answer 2

You have to find a parametrization of $C$ first: $$ \gamma(t)=\begin{cases} (t,0) & \mbox{ if } 0\le t \le 1\\ (1,t-1)&\mbox{ if } 1<t\le 2\\ (3-t,1)&\mbox{ if } 2<t\le 3\\ (0,4-t)&\mbox{ if } 3<t\le 4 \end{cases}. $$ With the above parametrization we have: \begin{eqnarray} \oint_C(y^2dx+x^2dy)&=&\sum_{k=0}^3\int_{k}^{k+1}(\gamma_2^2(t)\gamma_1'(t)+\gamma_1^2(t)\gamma_2'(t))\,dt=\int_1^2\,dt+\int_2^3-dt\\ &=&t\Big|_1^2-t\Big|_2^3=1-1=0. \end{eqnarray}