Help with $\int_{\Bbb D}(x^2 - y^2)\, dx\, dy$ where $\Bbb D=\{|x|+|y|\le2\}$

Question

Consider $\Bbb D=\{|x|+|y|\le2\}$; I'm trying to solve: $$\int_{\Bbb D}(x^2 - y^2)\, dx\, dy$$

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My goal is to solve that through a change of variables.
I was thinking to something like: $\begin{cases} |x|=u^2 \\ |y|=v^2 \end{cases}$. In such a case I would had the new domain as a circle; but the transformation is not invertible and I should study case by case: $\{x+y\le 2,x\ge0,y\ge0\},\{x-y\le2,x\ge0,y\le0\},\{-x+y\le2,x\le0, y\ge0\},\{-x-y\le2,x\le0,y\le0\}$.
Now, another transformation that would fit the problem is: $\begin{cases} x+y=u \\ x-y=v \end{cases}$, but I wasn't able to write any single domain in a comfortable way.

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Is there another suitable change of variables?
(otherwise) an someone handle one of the transformation I just proposed?

Answer 1

Note, the area under integration is a tilted square (with corners at (2,0), (0,2), (-2,0) and (0,-2). The sides of this square are given by the lines :$x-y=2, x-y =-2, x+y=2, x+y=-2$. Also, the integrand can be factored as: $x^2 - y^2 = (x-y)(x+y)$

Thus, choosing $u=x-y$ and $v=x+y$ seems a natural choice here.

Then, we use: $${\iint\limits_R {f\left( {x,y} \right)dxdy} } = {\iint\limits_S {f\left[ {x\left( {u,v} \right),y\left( {u,v} \right)} \right] }\kern0pt{ \left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right|dudv} ,}$$

where,$\left| {\large\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}\normalsize} \right|= \left| {\begin{array}{*{20}{c}} {\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}} \end{array}} \right|$

For this term, use $x=\frac{u+v}{2}$ and $y=\frac{v-u}{2}$. Getting, $\left| {\large\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}\normalsize} \right| = 1/2$

So we get, $I = \int_{u=-2}^2 \int_{v=-2}^2 \frac{uv}{2} du dv$ Which pretty easily evaluates out to $0$.

Answer 2

I highly recommend your second change of variables. For this choice of $u,v$, we find that $$ \mathbb D = \{x,y: -1 \leq u(x,y) \leq 1, -1 \leq v(x,y) \leq 1\}, $$ and we find that $$ x^2 - y^2 = u(x,y)v(x,y). $$

Answer 3

The domain is the square with vertices $(0,2),(-2,0),(0,-2),(2,0),$ and satisfies $$\begin{aligned}-x-2&\leq y \leq -x+2\\ x-2&\leq y \leq x+2\end{aligned}$$ From this $$-2\leq x+y \leq 2 \quad \text{and} \quad -2 \leq y-x \leq 2.$$

Answer 4

The integral is $0$ because the domain is symmethric relative to coordinate change $(x,y)\to (y,x)$ therefore $$\int_{\Bbb D}x^2\, \mathrm{d}x\, \mathrm{d}y= \int_{\Bbb D}y^2\, \mathrm{d}x\, \mathrm{d}y.$$