联立解 z = √(x^2+y^2), z = 8-x^2-y^2, 则 z > 0, 消去 z,
得两曲面交线在 xOy 坐标平面的投影是
x^2+y^2 = (17-√33)/2 < 8, x^2+y^2 = (17+√33)/2 > 8, 后者不合题设舍弃。
x^2+y^2 = (17-√33)/2 即 r = √[(17-√33)/2],
V = ∫∫
= ∫<0, 2π>dt ∫<0, √[(17-√33)/2]>(8-r^2-r)rdr
= ∫<0, 2π>dt ∫<0, √[(17-√33)/2]>(8r-r^2-r^3)dr
= 2π[4r^2-(1/3)r^3-(1/4)r^4]<0, √[(17-√33)/2]>
= 2π[r^2{4-(1/3)r-(1/4)r^2}]<0, √[(17-√33)/2]>
= π(17-√33){4-(1/3)√[(17-√33)/2]-(1/8)(17-√33)}
= π(17-√33){4-(1/3)√[(17-√33)/2]-(1/8)(17-√33)}
= (π/24)(17-√33){3(15+√33)-4√[(17-√33)/2]}
还可以这样吗?