Exercise: $\int_{\alpha}{( x^2 + y^2 )}d\alpha$ where $\alpha$ is the curve parameterized by $x=a(\cos(t)+ t\sin(t))$, $y=a(\sin(t)-t\cos(t))$ (correction by DonAntonio's) with $t \in [0,2\pi]$ where $a>0$
Proof
$M=\int_{\alpha}{( x^2 + y^2 )}d\alpha$
$M=\int_{0}^{2\pi}{a^2 \cos(t)^2 + a^2 t^2\sin(t)^2 + a^2 \sin(t)^2 + a^2 t^2\cos(t)^2}$ $dt$ ??? (it's that I have)
Solution
$M=16\pi^2(1+2\pi^2)$
Have some idea how to solve this integral of line, some hint to find the solution???
Example
If the circumference is $x^2+y^2=ax$ in the form $(x-a/2)^2 + y^2 = a^2/4$ where $C$ is parameterized by $x=(a/2)+(a/2)\cos(t)$ and $y=(a/2)\sin(t)$ with $(0\leq t\leq 2\pi)$
That mode
$\sigma'(t)=(-a\sin(t/2),a\cos(t/2))$ then $|\sigma'(t)|=a/2$
Then
$M=\int_{C}\sqrt{x^2+y^2}ds=\int_{0}^{2\pi}\sqrt{a^2/2 (1+\cos(t)a/2)}dt$
$...$
$M=\frac{a^2}{2}\cdot 2(1-\cos(t))^{1/2}|_{0}^{2\pi}=0$
I guess that there is a typo (see also DonAntonio's comment) and the parametrized curve $\alpha$ should be $$x:=a(\cos(t)+t\sin(t)),\quad y=a(\sin(t)-t\cos(t))$$ for $t\in [0,2\pi]$. Then $$x^2+y^2=a^2(1+t^2)\quad\mbox{and}\quad ds=\sqrt{(x'(t))^2+(y'(t))^2}=at.$$ Therefore $$\int_{\alpha}{( x^2 + y^2 )}ds=a^3\int_{0}^{2\pi}(1+t^2)t dt=2a^3\pi^2(1+2\pi^2).$$