Find the domain $D \subset \mathbb{R}^2$ such that the value of the integral $$ \iint\limits_D (1-x^2-y^2) \,\mathrm{d}A $$ is maximum.
I only have basic definitions and properties of Darboux integral on a general domain; no Fubini yet and no change of variables. I'm thinking to use the following facts:
- If $A \subset B$, $f$ is integrable on $A$ and $B$ with $f \ge 0$, then $\int\limits_A f \le \int\limits_B f$.
- If $f \le g$, then $\int\limits_D f \le \int\limits_D g$.
Applying to this problem, I'm going to find the maximum area of $D$ such that $1-x^2-y^2 \ge 0$. That means $D$ must be a disk center $O$ and radius $1$.
Am I right? I feel like my work is not concrete at all, not mathematically reasoning.
What should I do? Thank you very much.
Yes, the maximum value is attained for the domain $C:=\{(x,y): x^2+y^2\leq 1\}$, i.e. the disk centered at the origin of radius $1$. Indeed, if $D$ is any measurable set in $\mathbb{R}^2$ then $$\begin{align} \iint_D (1-x^2-y^2) \,\mathrm{d}A&=\iint_{D\cap C} (1-x^2-y^2) \,\mathrm{d}A +\iint_{D\setminus C} \underbrace{(1-x^2-y^2)}_{<0} \,\mathrm{d}A\\ &\leq \iint_{D\cap C} (1-x^2-y^2)\,\mathrm{d}A\\ &\leq \iint_{D\cap C} (1-x^2-y^2)\,\mathrm{d}A+\iint_{C\setminus D} \underbrace{(1-x^2-y^2)}_{\geq 0}\,\mathrm{d}A= \iint_{C} (1-x^2-y^2)\,\mathrm{d}A. \end{align}$$ Such maximum value can be easily calculated by using polar coordinates: $$\iint_{C} (1-x^2-y^2)\,\mathrm{d}A=2\pi\int_{0}^1(1-\rho^2)\rho\, d\rho=\frac{\pi}{2}.$$