Answer
$72$
Work Step by Step
$y=12-x^2$, $y=x^2-6$
Find their intersection points:
$12-x^2=x^2-6$
$2x^2=18$
$x^2=9$
$x=3, -3$
So the area is:
$\int_{-3}^3((12-x^2)-(x^2-6))dx$
$=\int_{-3}^3(18-2x^2)dx$
$=(18x-\frac{2x^3}{3})|_{-3}^3$
$=\left(18*3-\frac{2*3^3}{3}\right)-\left(18(-3)-\frac{2(-3)^3}{3}\right)$
$=(54-18)-(-54-(-18))$
$=36-(-36)$
$=72$