Let's find the integration boundaries:
−1≤x≤2 ;
to find the bounds for y equate y2+2 and 4−y2 :
y2+2=4−y2
2y2=2
y=±1
−1≤y≤1
y2+2≤z≤4−y2
V=∫−12dx∫−11dy∫y2+24−y2dz=∫−12dx∫−11(4−y2−(y2+2))dy=∫−12dx∫−11(2−2y2)dy=∫−12dx(2y−32y3)∣−11=38∫−12dx=38(2−(−1))=8
Answer: V=8.
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