First, find the coordinates of point A, and solve two equations simultaneously to get the coordinates of point A as (1√ (a+1), a/(a+ 1)). The integral area is [0, 1/√ (a+ 1)]. The equation of the straight line OA is y=ax/√(a+ 1), and the volume of the rotating body (i.e. the cone) obtained by the plane figure surrounded by the line segment OA, the X axis and the straight line X =1√ (a+1) rotating around the X axis. /√((a+ 1)？ √(a+ 1)).y=ax？ The plane figure surrounded by the X axis and the straight line X =1√ (a+1) rotates once around the X axis, and the volume of the rotator is v2 = ∫ (0,1√ (a+1)) π y? dx=π∫(0， 1/√(a+ 1))a？ x^4dx= 1/5*πa？ /((a+ 1)？ √(a+ 1)), and V 1-V2 is the number of revolutions. Therefore, V=V 1-V2=2/ 15*πa? /((a+ 1)？ √(a+ 1)), v' = π a (a+1) (-7/2) (4/15) Find the derivative of v so that V'=0 and the solution is a. Therefore, when a=4, the maximum value of V is 32π/(375√5).