Clarification on setting up the bounds of integration for $U = \{ (x,y) | -1 \leq x-y \leq 1 , \qquad1 \leq xy \leq 2 \}$

Question

I have a few challenges setting up the bounds of integration for the region $$U = \{(x,y) | -1 \leq x-y \leq 1 , \quad 1 \leq xy \leq 2 \}$$ My ultimate goal is to solve $$\iint_U x^2y + xy^2 dxdy = \iint_U f(x,y) dxdy$$

Here is a plot of the domain I have to consider:

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Since the region is symmetric, I decided to first focus on the first quadrant. The domain is not simple, therefore I divided it into simple domains by introducing two new bounds $x=1$ and $y=1$ yielding three simple domains: (Link to the interactive graph on Desmos)

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Starting with the region below $y=1$ then the middle region, then the upper one, I came up with the following limits:

$$\int_{ y= \frac{\sqrt5 - 1}{2} }^1 \int_{\frac1y}^{y+1} \space f(x,y) \space dxdy + \int_{ x=1 }^2 \int_{1}^{\frac2x} \space f(x,y) \space dydx + \int_{ y=1 }^2 \int_{\frac1y}^{1} \space f(x,y) \space dxdy $$

Now, the order turned out different for each of the summands, which is making me doubt that my setup is correct. Also, it seems to me that the points on the lines $x=1$ and $y=1$ will be ... counted twice (?) since they border two different integrals if that makes sense. Thus, could anyone kindly clarify on this.

Answer 1

Let's operat the substitution: $$u:=x-y\\ v=xy$$ EDIT: the transformation of the region was wrong Then, the region $U^+:=U \cap \lbrace (x,y): x>0, y>0\rbrace$ becomes $$ \tilde{U}:=\big\lbrace(u,v): -1\le u\le1 , 1\le v \le 2 \big\rbrace$$

Then, the Jacobian is: $$J= \begin{pmatrix} 1 & -1\\ y & x \end{pmatrix}$$ which has determinant $ |J|=x+y$. Then:

$$\iint_{U^+} x^2y + xy^2 dxdy = \iint_{U^+} xy(x + y) dxdy \iint_{\tilde{U}} xy\ (x + y) \frac{1}{x+y} dudv =\\ =\iint_{\tilde U} v\ dudv = \int_{1}^2\int_{-1}^1 v \ dudv = \int_{1}^2 2v \ dv =3$$

As @Gabrielek pointed out, the integral on the whole region is $0$, this is the velue on the upper right region $U^+$.

Answer 2

You've correctly written your domain, and you can easily see that is symmetrical with respect to the reflection $$r = \begin{cases} x \to -x \\ y \to -y \end{cases}$$

Let's define $U_+$ the region of you domain in the first quadrant and $U_-$ the other. Observe that $f(x,y) := x^2y + xy^2$ under the reflection written above behave like this $$f(r(x,y)) = f(-x,-y) = -x^2y - xy^2 = -f(x,y)$$

Now we can conclude because $$\begin{split} \iint_U f(x,y) dxdy &= \iint_{U_+} f(x,y) dxdy + \iint_{U_-} f(x,y) dxdy =\\ & =\iint_{U_+} f(x,y) dxdy + \iint_{U_+} -f(x,y) dxdy =\\ & = 0 \end{split}$$