Integration of a $2$-form What is
$$\int_C{\omega}$$
where $\omega=\frac{dx \wedge dy}{x^2+y^2}$ and $C(t_1,t_2)=(t_1+1)(\cos(2\pi t_2),\sin(2\pi t_2)) : I_2 \rightarrow \mathbb{R}^2 - \text{{(0,0)}}$?
The integrals of $1$-forms I understood as we take $dx$ and $dy$ as the derivatives of the parametrisation $C(t)$ w.r.t $t$. However for $2$-forms we now have the wedge product $dx \wedge dy$ which is where my problem lies in this question. Is this a matter of differentiation in two variables when computing this wedge? For example, before even calculating the wedge product, what is $dx(t_1,t_2)$?
The denominator isn't a problem: $x^2+y^2 = (t_1+1)^2$
 A: I've never seen a definition of the integral of a $2$-form along a "2D path" $C: I^2 \to \mathbb{R}^2$ (has anyone?), but it seems clear to me that the sensible definition should be
$$ \int_C \omega := \int_{I^2} C^*\omega$$
where $C^*\omega$ is the pull-back of $\omega$ by $C$. Recall that it is defined by
$$ (C^*\omega)_{|p} (u, v) := \omega_{|C(p)}(dC_{|p} (u), dC_{|p} (v))~.$$
NB: More generally, we could define the same way integrals $\int_C \omega$ where $\omega$ is a $k$-form on a $k$-dimensional manifold $M$ and $c$ is a smooth map $U \subset \mathbb{R}^k \to M$.
Let's come back to your problem. Practically, you can compute $C^* \omega$ by letting $(x,y) = C(t_1,t_2)$ in the expression of $\omega$, you get:
$$
\begin{align*}
x &= (t_1 + 1) \cos (2\pi t_2)\\
y &= (t_1 + 1) \sin (2\pi t_2) \\
\end{align*}
$$
hence
$$
\begin{align*}
x^2 + y^2 &= (t_1+1)^2\\
dx &= \cos (2\pi t_2)\,dt_1 - 2\pi (t_1 + 1)\sin(2\pi t_2) dt_2\\
dy &= \sin (2\pi t_2)\,dt_1 + 2\pi (t_1 + 1)\cos(2\pi t_2) dt_2\\
dx \wedge dy &= 2\pi (t_1+1)\, dt_1\wedge dt_2\\
\end{align*}
$$
thus
$$
C^* \omega = {2\pi\, dt_1\wedge dt_2\over t_1 +1}
$$
You can now easily compute your integral: assuming $I = [0,1]$ (you don't say what I is)
$$
\int_C \omega = \int_{I^2}{2\pi\, dt_1\wedge dt_2\over t_1 +1} = \int_{I^2}{2\pi\, dt_1 dt_2\over t_1 +1}
$$
which gives us by Tonelli's theorem
$$
\int_C \omega = 2\pi\left(\int_{0}^1{dt_1\over t_1 +1}\right)\left(\int_{0}^1 dt_2\right) = 2\pi \log (2)
$$
NB: Another approach ("just for fun") would be to work in polar coordinates, I'll let you try to figure out how do that.
A: Changing to polar coordinates, let $t_1+1=r\in[1,2]$, and $2\pi t_2=\phi\in[0,2\pi]$ (full circle). The set $C$ is an annulus in the $x,y$ plane bounded by the two concentric circles $r=1,2$, with $x^2+y^2=(r\cos\phi)^2+(r\sin\phi)^2=r^2$ The surface element $dx\wedge dy = r dr d \phi$. The function depends only on $r$, so the integration over the angle gives us simply $2\pi$, and finally the result is
$$2\pi\int_1^2 \frac{rdr}{r^2}=2\pi\log2$$
