∫[-√2→√2] √(8-2y²) dy
=√2∫[-√2→√2] √(4-y²) dy
令y=2sinu,則√(4-y²)=2cosu,dy=2cosudu,u：-π/4→π/4
=√2∫[-π/4→π/4] 4cos²u du
利用奇偶對(duì)稱性
=8√2∫[0→π/4] cos²u du
=4√2∫[0→π/4] (1+cos2u) du
=4√2(u+(1/2)sin2u) |[0→π/4]
=4√2(π/4+1/2)
=√2π+2√2