Let $\alpha (s)$ , $s\in [0,L]$, be a smooth positively oriented regular Jodan curve which is arc-length parametrized. The curve $\beta(s)=\alpha (s) +\lambda n(s)$, where $\lambda$ is a positive constant and $n(s)$ is the normal vector, is called a parallel curve to $\alpha$.
I have already proved that:
1)$ \text{Length}(\beta(s))=\text{Length}(\alpha) -2\pi \lambda$
2)$K_{\beta}(s)=\frac{K{\alpha}(s)}{1-\lambda K{\alpha}(s)}$
Now, I have to compute the area of $\beta$ in terms of the area of $\alpha$: $A(\Omega_{\alpha})$
$\beta(s)=\alpha (s) +\lambda n(s)= (\alpha_1 (s) +\lambda n_1(s),\alpha_2 (s) +\lambda n_2(s))$
$\beta'(s)=\alpha '(s) +\lambda n'(s)=(\alpha _1'(s) +\lambda n_1'(s),\alpha _1'(s) +\lambda n'_1(s))$
By a corollary of Green's theorem,
$A(\Omega_{\beta})=-\int_{0}^L (\alpha_2 (s) +\lambda n_2(s))(\alpha _1'(s) +\lambda n_1'(s))ds=$$-\int_{0}^L (\alpha_2\alpha_1'+\lambda \alpha_2 n_1'+\lambda n_2\alpha _1'+\lambda^2n_2n_1') $=
Using, Frenet formulas: $n_1'(s)=-k(s)\alpha_1'(s)$ we got:
=$A(\Omega_\alpha)+\int_{0}^L \lambda k\alpha_2\alpha_1'-\int_{0}^L n_2\alpha_1'+\int_{0}^L \lambda^2kn_2\alpha_1'$
I don't know how to continue or if this is the right way to do it.
***Are the previous formulas for the curvature length and area still true for a general $\alpha$?
Thank you for your help.
Let $\gamma_t(s)=\alpha(s)+t \mathbb{n}, \forall t\in [0,\lambda]$. We have $\gamma_0=\alpha$ and $\gamma_{\lambda}=\beta$, and the area between $\alpha$ and $\beta$ is parameterized by $(s,t)$. Denote $k(s)$ as the curvature of $\alpha$ at $s$. First let's show that the tangent vector using the notation from wiki and using outer normal as the positive direction: $$\frac{\partial\gamma}{\partial t}=\mathbb{n}$$ $$\frac{\partial\gamma}{\partial s}=\dot{\alpha} +t\dot {\mathbb{n}}=T+tkT=(1+tk)T $$ Since $\{T, \mathbb{n}\}$ is an orthonormal basis the area element can be written as $$d\sigma=(1+tk)dt\wedge ds$$ Hence $$A(\Omega_\beta)-A(\Omega_\alpha)=\int d\sigma$$ $$=\int(1+t\cdot k)dt\wedge ds$$ $$=\int\int dt ds + \int t\cdot ({\int kds}) dt$$ $$=\lambda L + 2\pi\int t dt$$ $$=\lambda L + \pi \lambda^2$$