Line integrals in differential form 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!)?
 A: 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
$$\bbox[5px,border:2px solid #C0A000]{\oint_C (y^2dx+x^2dy)=0}$$
A: 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}
