Integration for finding the Arc Length of Circle $x^2+y^2=a^2$ Question:

Find the arc length of the circle given by $x^2+y^2=a^2$.   
$Ans = 2\pi a$

How to obtain the ans? I have no ideas after doing the following thing.

Thank you for your attention.

For curve $\gamma$ through $\mathbb{R}^n$, i.e. for a function $[a,b] \,:\, \mathbb{R} \to \mathbb{R}^n$ the arc length of $\gamma$ is $$
  L(\gamma) = \int_a^b \| f'(t) \| \,dt \quad\text{where $f'$ is the gradient of $f$, i.e. $f'(t) = \left(\frac{\partial f}{\partial x_1},\ldots,\frac{\partial f}{\partial x_n}\right)$.}
$$
The function $f(t) = \left(a\cos t, a\sin t\right)$ for $t \in [0,2\pi]$ is a parametric description of the circle $x^2 + y^2 = a^2$. You have $$
  \|f'(t)\| = \left\|\left(-a\sin t, a\cos t\right)\right\| = \sqrt{a^2\sin^2 t + a^2\cos^2 t} = a\sqrt{\sin^2 t + \cos^2 t} = a \text{,}
$$
and thus $$
  L(f) = \int_0^{2\pi} \|f'(t)\| \,dt = \int_0^{2\pi} a \,dt = a(2\pi - 0) = 2\pi a \text{.}
$$
