Question regarding curl in dimensions higher than 3 According to the wikipedia page about curl curl can be defined implicily
as 
$$(\nabla \times \textbf{F} ) \scriptsize{\bullet} \normalsize{\hat{n}} = \lim_{A \rightarrow 0} \frac{1}{|A|} \oint_C \textbf{F} \scriptsize{\bullet} \normalsize{\textbf{dr}}$$
My question is, specifically, is this definition limited to only two dimensions? If not, how would I apply it to higher dimensions?
For example, if I had a vector field defined by $\textbf{F}(x,y)=<x^2-y,y^2+x>$ the line integral at (a,b) could be defined by 
$$x-a=r \cos(t) \wedge y-b=r \sin(t)$$
$$\frac{dx}{dt}=-r\sin(t) \wedge \frac{dy}{dt}=r\cos(t)$$
$$\oint_C \textbf{F} \scriptsize \bullet \normalsize \textbf{dr} \\= \int_0^{2\pi} \large [ \normalsize (r \ \cos(t)+a)^2-(r \ \sin(t)+b) \large] \normalsize \frac{dx}{dt} dt + \int_0^{2\pi} \large [ \normalsize (r \ \sin(t) + b)^2+(r \cos(t) + a ) \large ] \normalsize \frac{dy}{dt} dt\\= 2 \pi r^2$$
and $A = \pi r^2$ so $\frac{1}{A}=\frac{1}{\pi r^2}$ which makes
$$\lim_{A \rightarrow 0} \frac{1}{A} \oint_C \textbf{F} \scriptsize{\bullet} \normalsize{\textbf{dr}} = \lim_{r \rightarrow 0} \frac{2 \pi r^2}{\pi r^2} =\lim_{r \rightarrow 0} 2 = 2 $$
Since $\hat{n}$ for the $xy$-plane is simply $<0,0,1>$ and since the curl would also be pointing in the same direction, then the curl, expressed as a vector in 3-space must be $<0,0,2>$. On WolframAlpha, entering curl [x^2-y,y^2+x,0] gives us the exact same answer.
That was a little involved, to be honest, but I feel it was necessary for this question. A follow up question I have to ask is, if I were to apply this specifically to three dimensions, would a simple line integral in 3-space provide a sufficient answer, or would it need a surface integral?
 A: Instead of some analysis, let's take a step back and think about geometry. What's going on when you take the cross product of two vectors? You're defining a plane and then using the plane's normal field to find a vector.
Informally, the curl is the del operator cross-product with a vector field: we write $\operatorname{curl}X = \nabla\times X$ for a reason. So what's happening geometrically? The curl of a vector field measures infinitesimal rotation. Rotations happen in a plane! The plane has a normal vector, and that's where we get the resulting vector field. So we have the following operation:
$$ \mbox{vector field}\to \mbox{planes of rotation}\to \mbox{normal vector field}  $$
This two-step procedure relies critically on having three dimensions. In higher dimensions, a plane doesn't have just one normal vector, it has many normal vectors. So, unfortunately, we can't use this "measure the plane of infinitesimal rotation and then take a normal field" business to produce a vector field.
This is because the curl is really the exterior derivative in disguise. Given a vector field, there is a one-form dual to it. The exterior derivative produces a two-form from that one-form (this is the same as our weighted plane field), and Hodge duality uniquely identifies another one-form, which is dual to the normal vector field. Here's a brief description.
