Exercise 3.4.10

• If $u(x,y)=x+y$, we have $\int {\int \; <!--\; nolimits\; -->}_{x,y\text{ in triangle}}\parallel \mathit{gradu}{\text{∥}}_2\mathit{dxdy}=\int \; <!--\; nolimits\; -->{\int \; <!--\; nolimits\; -->}_{x,y\text{ in triangle}}\parallel (1,1){\text{∥}}_2\mathit{dxdy}=\int \; <!--\; nolimits\; -->{\int \; <!--\; nolimits\; -->}_{x,y\text{ in triangle}}\sqrt{2}\mathit{dxdy}=\sqrt{2}\int \; <!--\; nolimits\; -->{\int \; <!--\; nolimits\; -->}_{x,y\text{ in triangle}}\mathit{dxdy}=\frac{\sqrt{2}}{2}$

Using the right hand side, we have if $u(x,y)=t$, which is a line segment within the triangle. It has length of $\sqrt{t^2+t^2}=\sqrt{2}t$ within the triangle. So ${\int }_0^1\sqrt{2}\mathit{tdt}=\sqrt{2}{\int \; <!--\; nolimits\; -->}_0^1\mathit{tdt}=\frac{\sqrt{2}}{2}$

• If $u=0$ in a unit square and $u=1$ outside, $t$ can only be 0 or 1. If $t=0$, the length is 4, the length of the unit square. if $t=1$, $u=1$ is a plane with middle part missing. its length is infity because the $u$ here has domain on the whole $x,y$ plane. So $\parallel u{\text{∥}}_{\mathit{BV}}=\infty$. Not sure if this is correct though?