The $l^p$ norms $\mid(x,y)\mid_p \lvert(x,y)\rvert_p = (\mid x\mid^p+\mid \lvert x\rvert^p+\lvert y \mid^p)^{1/p}$ rvert^p)^{1/p}$are norms and satisfies that if$\mid(x,y)\mid_p=1$\lvert(x,y)\rvert_p=1$ and $q>p$ then $\mid(x,y)\mid_q\leq \lvert(x,y)\rvert_q\leq 1$. So the unit "circles" of which you want to find the area grows.
It is also a fact that $\mid \lvert (x,y) \mid_p rvert_p \to \max (\mid x\mid,\mid y\mid)$ \lvert x\rvert,\lvert y\rvert)$as$p\to \infty$. So the unit circles converges to the square which is the boundary of$[-1,1]\times [-1,1]$. This implies by monotone convergen convergence theorem that your integral converges to 1. Because the entire square has area 4. 1 I think this question smells of homework, but another answer, which to me totally obscures the geometric nature of the question has been posted, and I feel that this justifies the following answer (even if the question is closed): The$l^p$norms$\mid(x,y)\mid_p = (\mid x\mid^p+\mid y \mid^p)^{1/p}$are norms and satisfies that if$\mid(x,y)\mid_p=1$and$q>p$then$\mid(x,y)\mid_q\leq 1$. So the unit "circles" of which you want to find the area grows. It is also a fact that$\mid (x,y) \mid_p \to \max (\mid x\mid,\mid y\mid)$as$p\to \infty$. So the unit circles converges to the square which is the boundary of$[-1,1]\times [-1,1]\$. This implies by monotone convergen theorem that your integral converges to 1. Because the entire square has area 4.