旋轉體體積(10分)

Since it is not revolving about the y-axis but about x = -1, we first translate the function to the right by 1 unit, so the question is changed to find the volume when y = sqrt (x -1) and x = 2 is revolved about the y -axis. So volume
= S p[ 2^2 - x^2] dy from y = -1 to y = 1. where p = pi
= p S [4 - (y^2 + 1)^2] dy
= p S [ 4 - y^4 - 1 - 2y^2) dy
= p[3y - y^5/5 - 6y^3/3] from y= -1 to y = 1
= p(3 - 1/5 - 2 + 3 - 1/5 - 2)
= 8p/5.

2008-11-25 08:21:36 補充：
Correction: Since the area is not from y = - 1 but from y = 0, so the volume should be half of 8p/5 = 4p/5.

2008-11-25 08:27:54 補充：
Correction: Line 7 should be p[3y - y^5/5 - 2y^3/3] from y = 0 to y = 1. That is p(3 - 1/5 - 2/3) =32p/15. (not 4p/5, sorry).