Given before question: $\hat{\mathbb{X}}_t\circ\hat\Phi_t=\frac{\partial}{\partial t}\hat\Phi_t$ and $$\beta=\frac{(x^2-y^2)dy\wedge dz+2xydz\wedge dx}{(x^2+y^2)^2}$$ I have shown that $d\beta=0$
Question: Let $\hat\Phi_t:U\rightarrow U$ be given by $\hat\Phi_t(x,y,z)=(tx,y,z)$
(part 1)Find $\hat{\mathbb{X}}_t$ as defined above, and (part 2)find a one-form $\alpha_0$ such that
$$\lim_{t\to0}\hat\Phi_t^*\beta=d\alpha_0$$
Answer part $1$: I have found $\hat{\mathbb{X}}_t$ as follows:$\hat{\mathbb{X}}_t(x,y,z)=((\frac{\partial}{\partial t}\hat\Phi_t)\circ(\hat\Phi_t^{-1}))(x,y,z)=\frac{\partial}{\partial t}\hat\Phi_t(\frac{x}{t},y,z)=\frac{\partial}{\partial t}(x,y,z)=(x',y',z')$
Answer part $2$: $$\lim_{t\to0}\hat\Phi_t^*\beta=d\alpha_0\Leftrightarrow \lim_{t\to0}\beta\circ\hat\Phi_t=d\alpha_0$$
$$\lim_{t\to0}(\beta\circ\hat\Phi_t)(x,y,z)=\lim_{t\to0}\beta(tx,y,z)$$
Since the $x$ component will always be zero in $\beta$ we get:
$$\lim_{t\to0}\beta(tx,y,z)=\lim_{t\to0}\beta(0,y,z)=\frac{-y^2dy\wedge dz}{y^4}=-\frac{1}{y^2}dy\wedge dz$$
Therefore $d\alpha_0=-\frac{1}{y^2}dy\wedge dz$
Point of contention: How do I compute this last result? What is $\alpha_0$ if $d\alpha_0=-\frac{1}{y^2}dy\wedge dz$?
I have never found the integral of a wedge product, how should I do this part? I may be able to work it out through trial and error, but I would prefer a method to use. Are there any standard results to know for computing integrals of this kind (Integral of $2$-forms)?
Extra: Just realised that I did not take into account the limit, how does this affect the answer? Would we compute $d\alpha_0=-\lim_{t\to0}\frac{1}{y^2}dy\wedge dz$ or $d\alpha_0=-\frac{1}{y_0^2}dy_0\wedge dz_0$ or what else. I am quite clueless as for this last part.
By definition of $\Bbb X_t$, $\Bbb X_t(\Phi_t(x,y,z)) = \frac{\partial}{\partial t}(\Phi_t(x,y,z))$, i.e., $\Bbb X_t(tx,y,z) = \frac{\partial}{\partial t}(tx,y,z)$. Thus $\Bbb X_t(tx,y,z) = (x,0,0)$, and consequently $\Bbb X_t(x,y,z) = (\frac{x}{t},0,0)$. Since $$\Phi_t^*\beta = \frac{t^2x^2 - y^2}{(t^2x^2 + y^2)^2}\, dy\wedge dz - \frac{2txy}{(t^2x^2 + y^2)} t\,dx\wedge dz$$ we have $$\lim_{t \to 0} \Phi_t^*\beta = -\frac{y^2}{y^4}\, dy\wedge dz = -\frac{1}{y^2}\, dy\wedge dz = d\left(\frac{1}{y}\right)\wedge dz = d\left(\frac{1}{y}\, dz\right) = d\alpha_0,$$ where $$\alpha_0 = \frac{dz}{y}.$$