Verify the relation: 0 < double integral over D xy(x+y)

Verify the relation: $$ 0 < \iint_D xy(x+y)\,dx\,dy < 64, $$ where $D=\{(x,y): 0<x<2,\ 0<y<2\}$.

Step 1 · Identify the region

$D=\{(x,y):0<x<2,\,0<y<2\}$; for integration purposes we use the closed square $[0,2]\times[0,2]$, with area $A=4$.

Step 2 · Analyze the function

$$ f(x,y)=xy(x+y)=x^2y+xy^2. $$

Since $x,y\geqslant0$ on $D$, we have $f(x,y)\geqslant0$.

Step 3 · Find the minimum of $f$

$f(0,y)=0$ and $f(x,0)=0$ on the boundary where $x=0$ or $y=0$, so $\min_{(x,y)\in D} f(x,y)=0$.

Step 4 · Find the maximum of $f$

$$ \frac{\partial f}{\partial x}=y(2x+y), \qquad \frac{\partial f}{\partial y}=x(x+2y). $$

Setting both to zero with $x,y>0$ forces $y=-2x$ and $x=-2y$ simultaneously, which only holds at $x=y=0$ — on the boundary, not the interior. So the maximum occurs on the boundary of $D$.

Step 5 · Check the boundary

Edge $x=0$ and edge $y=0$ give $f\equiv0$. Edge $x=2$: $f(2,y)=4y+2y^2$, maximized at $y=2$: $f(2,2)=16$. Edge $y=2$: $f(x,2)=2x^2+4x$, maximized at $x=2$: $f(2,2)=16$. Hence

$$ \max_{(x,y)\in D} f(x,y)=f(2,2)=2\cdot2\cdot4=16. $$

Step 6 · Apply the estimation theorem

$$ 0\cdot4 \leqslant \iint_D xy(x+y)\,dx\,dy \leqslant 16\cdot4 \quad\Longrightarrow\quad 0\leqslant I\leqslant 64. $$

Since $f>0$ in the interior of $D$, the integral is strictly positive: $0<I\leqslant64$.

Step 7 · Calculate the exact integral

$$ \iint_D(x^2y+xy^2)\,dx\,dy = \int_0^2\int_0^2 x^2y\,dx\,dy + \int_0^2\int_0^2 xy^2\,dx\,dy. $$

$$ \int_0^2\int_0^2 x^2y\,dx\,dy = \left[\frac{y^2}{2}\right]_0^2\left[\frac{x^3}{3}\right]_0^2 = 2\cdot\frac83=\frac{16}{3}, \qquad \int_0^2\int_0^2 xy^2\,dx\,dy = \frac{16}{3}. $$

$$ \iint_D xy(x+y)\,dx\,dy = \frac{16}{3}+\frac{16}{3}=\frac{32}{3}\approx10.667. $$

Step 8 · More precise bounds

Since the exact value is $\frac{32}{3}\approx10.667$, the bound $I\leqslant64$ is loose but valid:

$$ 0 < \frac{32}{3} < 64. \quad \checkmark $$

Summary

Key steps: on $[0,2]\times[0,2]$, $f(x,y)=xy(x+y)$ has $\min f=0$ (on the edges $x=0$ or $y=0$) and $\max f=16$ (at the corner $(2,2)$); multiplying by the area $4$ gives $0<I\leqslant64$; the exact value $I=\frac{32}{3}$ confirms the inequality.

Additional Notes

• The bound $64$ is conservative — the actual value $\frac{32}{3}$ is much closer to $0$, since $f$ is small over most of the square and only large near the corner $(2,2)$.
• Separating $x^2y+xy^2$ into two products of single-variable factors makes exact evaluation straightforward.