Consider a function $f: \Omega \to \mathbb{R}$, which $\Omega \subset \mathbb{R}^n$ is an open set. Then its total variation is defined as $V(f, \Omega )= \sup \{ \int_{\Omega} f(x) ~ \mathrm {div}g(x)~ dx : g \in C_c^1 (\Omega), \|g\|_{L^\infty } \leq 1 \}$.
I want to calculate the total variation for the function $f:\mathbb{R}^2 \to \mathbb{R}$ such that $f(x,y)=1$ for $x^2+y^2 \leq 1$; $f(x,y)=2-\sqrt{x^2+y^2}$ for $1 \leq x^2+y^2 \leq 2$; and $f(x,y)=0$ otherwise. (Another question is the funtion $g= \chi_D$, the charectristic function of the unit ball in $\mathbb{R}^2$, which I think its total variation is equal to the priemeter of the unit ball, which I have no proof). In fact I want to know is there motivation to estimate total variation whitout direct using of the formula?
First let us study the case of $f = \chi_{B_r},$ where $B_r = \{ (x,y) \in \mathbb R^2 \mid x^2+y^2 \leq r^2 \}.$ The integral under supremum then becomes $$ \int_{\mathbb R^2} \chi_{B_r}(x,y) \, \nabla\cdot g(x,y) \, dx \, dy = \int_{B_r} \nabla\cdot g(x,y) \, dx \, dy = \oint_{\partial B_r} g(x,y) \cdot n(x,y) \, dl(x,y) $$ which is maximized (under the restriction $\|g\|_{L^\infty}\leq 1$) when $g$ is equal to $n$ on $\partial B_r.$ The value then becomes equal to the length of $\partial B_r,$ i.e. $2\pi \, r.$
Now, lets look at $f$ defined by $$f(x,y) = \begin{cases} 1 & \text{ when } x^2+y^2 \leq 1, \\ 2-\sqrt{x^2+y^2} & \text{ when } 1 \leq x^2+y^2 \leq 2, \\ 0 & \text{ when } x^2+y^2 > 2. \end{cases}$$
The integral under supremum can be divided into three parts: $$ \int_{\mathbb R^2} f(x,y) \, \nabla \cdot g(x,y) \, dx \, dy = \int_{x^2+y^2 \leq 1} 1 \, \nabla \cdot g(x,y) \, dx \, dy \\ + \int_{1 \leq x^2+y^2 \leq 2} (2-\sqrt{x^2+y^2}) \, \nabla \cdot g(x,y) \, dx \, dy \\ + \int_{2 \leq x^2+y^2} 0 \, \nabla \cdot g(x,y) \, dx \, dy. $$
The first and second parts can be partially integrated giving $$ \int_{x^2+y^2 \leq 1} \nabla \cdot g(x,y) \, dx \, dy = \oint_{x^2+y^2 = 1} g(x,y) \cdot n(x,y) \, dl(x,y) $$ and $$ \int_{1 \leq x^2+y^2 \leq 2} (2-\sqrt{x^2+y^2}) \, \nabla \cdot g(x,y) \, dx \, dy \\ = \int_{1 \leq x^2+y^2 \leq 2} \nabla \cdot \left( (2-\sqrt{x^2+y^2}) \, g(x,y) \right) \, dx \, dy \\ - \int_{1 \leq x^2+y^2 \leq 2} \nabla \left( 2-\sqrt{x^2+y^2} \right) \cdot g(x,y) \, dx \, dy \\ = \int_{x^2+y^2 = 2} (2-\sqrt{x^2+y^2}) \, g(x,y) \cdot n(x,y) \, dl(x,y) \\ - \int_{x^2+y^2 = 1} (2-\sqrt{x^2+y^2}) \, g(x,y) \cdot n(x,y) \, dl(x,y) \\ - \int_{1 \leq x^2+y^2 \leq 2} \nabla \left( 2-\sqrt{x^2+y^2} \right) \cdot g(x,y) \, dx \, dy \\ = \int_{x^2+y^2 = 2} (2-\sqrt{2}) \, g(x,y) \cdot n(x,y) \, dl(x,y) \\ - \int_{x^2+y^2 = 1} g(x,y) \cdot n(x,y) \, dl(x,y) \\ - \int_{1 \leq x^2+y^2 \leq 2} \nabla \left( 2-\sqrt{x^2+y^2} \right) \cdot g(x,y) \, dx \, dy $$ The third part obviously vanishes.
When summing the parts, we see that the boundary integrals at $x^2+y^2=1$ cancel and we end up with $$ \int_{x^2+y^2 = 2} (2-\sqrt{2}) \, g(x,y) \cdot n(x,y) \, dl(x,y) \\ - \int_{1 \leq x^2+y^2 \leq 2} \nabla \left( 2-\sqrt{x^2+y^2} \right) \cdot g(x,y) \, dx \, dy \\ = \int_{x^2+y^2 = 2} (2-\sqrt{2}) \, g(x,y) \cdot n(x,y) \, dl(x,y) \\ + \int_{1 \leq x^2+y^2 \leq 2} \hat r \cdot g(x,y) \, dx \, dy, $$ where $\hat r(x,y) = (x \hat x + y \hat y)/\sqrt{x^2+y^2}.$
Both these integrals are maximized if we take $g(x,y) = \hat r(x,y)$ (note that $n(x,y) = \hat r(x,y)$ on $x^2+y^2=2$), which gives the value $$ \int_{x^2+y^2 = 2} (2-\sqrt{2}) \, dl(x,y) + \int_{1 \leq x^2+y^2 \leq 2} dx \, dy = (2-\sqrt{2}) \, 2\pi \, \sqrt{2} + \pi (2^2 - 1^2) = (4 \sqrt 2 - 1) \pi. $$
Make sure that you understand what I have done. I usually do some mistake somewhere.