y=x^2

y=2-x

x^2=2-x

x^2+x-2=0

(x+2)(x-1) = 0

x= -2, 1 (limits of integration)

Volume = pi ∫ ((4-x^2)^2 - (4-(2-x))^2 ) dx

= pi ∫ (x^4-9x^2-4x+12) dx

= 36pi/5

use [ π ( large radius² - small radius² ) thickness ]...large is [ 4 - x² ] , small is [ 4 - ( 2 - x)] for x in [ 0 , 1 ] , thickness is [ dx ]....

use [ 2π radius height thickness ]...radius is [ 4 - y ] , height is [ ( 2 - y ) - 0 ] when y in [ 1, 2 ] & [ √y - 0 ] when y in [ 0 , 1 ] ;

thickness is [ dy ] , thus 2 integrals required ...FINAL comment : your description yields 2 bounded regions , I used x ≥ 0