Suppose $U:={\iint}_{R} (x^2 + 2y^2+9) \,dx\,dy$ and $V:= \iint _R (2x^2 + 3y^2)\, dx\,dy$. Determine the integration region $R$ where $U \geq V$. Hence, find the value $U-V$ over this region.
Attempt:
Since $U \geq V$, we have $\iint _R(-x^2 - y^2 +9) \geq 0$. Then is it correct to say that $-x^2 -y^2 +9 \geq 0$?
We want to achieve $$U:=\iint_R x^2+2y^2 + 9 \, dx \, dy \ge \iint_R 2x^2+3y^2 \, dx \, dy=:V.$$ So $$\iint_R -(x^2+y^2)+9 \, dx \, dy \ge 0,$$ as you stated in your original post. For the double-integral to be positive, we should have $$-(x^2+y^2)+9 \ge 0,$$ like you said. Therefore, $$x^2+y^2 \le 9.$$ This is the equation of a circle of radius $3$, centered at $(0,0)$. And $U \ge V$ over all $(x,y)$ in this circle.
