作OP ⊥ 平面ABCD, 交平面ABCD于P, 连PA, PB, PC, PD.
由A, B, C, D在以O为球心半径为4的球面上, 有OA = OB = OC = OD = 4.
于是由勾股定理可得PA = PB = PC = PD,
即P是矩形ABCD外接圆圆心, 也即对角线AC与BD的交点和中点.
AC = √(AB²+BC²) = √39, PA = AC/2 = (√39)/2.
OP = √(OA²-PA²) = √(25/4) = 5/2.
S_ABCD = AB·BC = 6√3.
故V_O-ABCD = 1/3·S_ABCD·OP = 1/3·6√3·5/2 = 5√3.