Compute $\int_Rx^2+y^2$, where $R=\{(x,y)\in\mathbb{R}^2:|x|\leq|y|\leq2\}$.

Question

Compute the integral of $f(x,y)=x^2+y^2$ over $R$, where $R=\{(x,y)\in\mathbb{R}^2:|x|\leq|y|\leq2\}$.

Writing the region as $R=\{(x,y)\in\mathbb{R}^2:-2\leq y\leq2, -y\leq x\leq y\}$, we have that:

$$\int_Rx^2+y^2=\int_{-2}^2\int_{-y}^yx^2+y^2dxdy=0$$

If we think that $R$ can be divided into two pieces $R_1=\{(x,y)\in\mathbb{R}^2:-2\leq y\leq x\leq0\}$ and $R_2=\{(x,y)\in\mathbb{R}^2:0\leq x\leq y\leq2\}$, then $\int_{R_1}f=-\int_{R_2}f$, and therefore $\int_Rf=0$.

Is my answer right? Am I making any wrong assumption? Thanks in advance!

Answer 1

By symmetry in $x$ and $y$ the integral is the same as that over $R'$ defined by $|y|\le|x|\le2$. Adding the two integrals gives $$\int_{-2}^2\int_{-2}^2(x^2+y^2)\,dx\,dy$$ and your integral is a half of that.

Answer 2

I view of the symmetry of the integrand in $x$ and $y$ the integral is $4\int_W (x^{2}+y^{2}) d(x,y)$ where $W=\{(x,y): =0\leq x \leq y \leq 2\}$. You can write this as $\int_0^{2}\int_x^{2}(x^{2}+y^{2}) dydx=\frac {64} 3$.