## Thomas' Calculus 13th Edition

$x^2+2y^2 \leq 4$ This is the region inside the ellipse $x^2+2y^2=4$
We have $\iint_{R} (4-x^2-2y^2) dA$ The integral must be a maximum on the region when $4-x^2-2y^2\gt 0$ We will have to multiply with $-1$, a negative number, so that the direction of the inequality reverses. So, $4-x^2-2y^2\leq 0$ or, $x^2+2y^2 \leq 4$ This is the region inside the ellipse $x^2+2y^2=4$