Find the area of the region between the curves y=8−x^2 y=x^2 x=−3 and x=3

I made a slight correction

symmetric around x = 0 so do it from x = 0 to x = 3, then double.

crash into each other at x = 2
so
do it from x = 0 to x = 2
then from x = 2 to x = 3
add, then double

integral (8-2x^2)dx from 0 to 2
8x - (2/3)x^3 at 2 - at 0
16 -16/3

integral from 2 to 3
(2 x^2 - 8) dx
(2/3)x^3 - 8x at 3 - at 2
18-24 -(16/3-16)
10 -16/3

now add
26 -32/3
double
52 - 64/3
check my arithmetic !