–√2 ≤ x ≤ √2
0 ≤ y ≤ 2–x2
∫ ∫D(x+y)dxdy=∫^(√2–√2 (∫2–x20(x+y)dy )dx=
=∫^(√2–√2 (xy+(y2/2) )|2–x20dx=
=∫^(√2–√2 (x·(2–x2)+((2–x2)2/2) )dx=
=∫^(√2–√2 (2x–x3+(4–4x2+x4)/2)dx=
=((2x2/2)– (x4/4) +2x–(2x3/3)+(x5/10))|^(√2–√2=
=(2–2) –(1/4) (4–4)+2·(√2–(–√2))–(2/3)·(2√2+2√2)+(1/10)·(4√2+4√2)=
=4√2–8√2/3 +8√2/10= 32√2/15