數學求積法一問(2題)

(A) When the cylinder is lying horizontally, let the side of the water surface be AB, let center of circle be O.
OA = radius of circle = 6
Let mid - point of AB be M, so OM = 6 - 4 = 2
By Pythagoras thm., AM = sqrt ( 36 - 4) = sqrt 32 ( AB = 2 sqrt 32)
So area of triangle OAB = [(6 - 4) 2sqrt 32]/2 = 2 sqrt 32 = 11.313708....(1)
Angle AOM = arccos[(6 - 4)/6] = arccos(1/3) = 70. 5288, so angle AOB = 2 x 70. 5288 = 141.0576
So area of sector OAB = pi x 36 x ( 141.0576/360) = 44.314655...(2)
From (1) and (2), area of cross - section of water = (2) - (1) = 33
So volume of water = 33 x 30 = 990.
Let height of water = h
So pi x 36 x h = 990
h = 990/(pi x 36) = 8.7535 = 8.75.


2012-11-19 08:29:12 補充：
(B) Volume of bowl = volume of hemisphere = (2/3) x pi x 5^3 = 250pi/3.
Let height of the larger cone = h,so by similar triangle, h/9 = (h - 8)/5
5h = 9h - 72
h = 72/4 = 18
so height of smaller cone = 18 - 8 = 10
So volume of wood = (1/3)pi[ 9^2 x 18 - 5^2 x 10] - 250pi/3 = 958pi/3.