calculus - Show $int_{gamma}F cdot N ds = int_{gamma} omega $

So I am stuck on this problem and would really appreciate some help.

Ok here is the question:

Let $\gamma$ be a smooth parameterization of the curve $\zeta$ in $R^2$
, with unit tanget $T$ and unit normal $N$ defined by $$T(\gamma(t)) = \frac{\gamma'(t)}{\left|\gamma'(t)\right|}=\frac{1}{\left|\gamma'(t)\right|}(\gamma_1'(t),\gamma_2(t))$$

and

$$N(\gamma(t)) = \frac{1}{\left|\gamma'(t)\right|}(\gamma_2'(t),-\gamma_1'(t))$$

Given a vector field $F=(f_1,f_2)$ on $R^2$, let $\omega = -f_2dx + f_1dy$, and then show that $$\int_{\gamma}F \cdot N ds = \int_{\gamma} \omega $$

thanks guys!

Answer

Notice $\omega$ is an infinitesimal rotated version of $F$. First we rewrite the integral as inner product
$$
\int_\gamma\omega = \int_\gamma-f_2dx + f_1dy = \int_\gamma (-f_2,f_1)\cdot (dx,dy)
$$
Plugging the parametrization $x=\gamma_1(t), y= \gamma_2(t)$, and switching positions for the inner product:
$$\begin{aligned}
&\int_\gamma (-f_2,f_1)\cdot (dx,dy) = \int_a^b (-f_2,f_1)\cdot (\gamma_1'(t),\gamma_2'(t))dt

\\
=& \int_a^b (f_1,f_2)\cdot (\gamma_2'(t),-\gamma_1'(t))dt
\\
=& \int_a^b (f_1,f_2)\cdot \frac{(\gamma_2'(t),-\gamma_1'(t))}{|\gamma'(t)|} |\gamma'(t)|dt
\\
=& \int^b_a F\cdot N |\gamma'(t)|dt = \int_\gamma F\cdot N ds
\end{aligned}
$$
The last step is just by the definition of the line integral of a scalar function.