∵x²=a²-az,x²+y²=（a／2）²,z=0(a＞0)
∴所圍立體是一個(gè)下底以a/2為半徑的圓,上底以曲面az=y²+3a²/4為頂?shù)膱A柱體
故所圍立體的體積=4∫(0,a/2)dx∫(0,√(a²/4-x²))[y²/a+3a/4]dy (∫(0,a/2)表示從0到a/2積分,其他類同)
=4∫(0,π/2)dθ∫(0,a/2)(r²sin²θ/a+3a/4)rdr (進(jìn)行極坐標(biāo)變換)
=4∫(0,π/2)dθ∫(0,a/2)(r³sin²θ/a+3ar/4)dr
=∫(0,π/2)dθ*(r^4*sin²θ/a+3ar²/2)│(0,a/2)
=(a³/16)∫(0,π/2)(sin²θ+6)dθ
=(a³/32)∫(0,π/2)(13-cos(2θ))dθ
=(a³/32)(13θ-sin(2θ))│(0,π/2)
=(a³/32)(13π/2)
=13πa³/64.