Question 2177
ok, got it :-)

Let centre of circle be O, Let height of balls be a horizontal line AB, which is 1.14m above the bottom of the cyclinder (and is above the centre of the circle) since cylinder radius = 1.105m.

I shall consider the free space above the balls...this is the segment made by the line AB, as part of the sector AOB. Given that the angle AOB is a (in radians!!), then we know the following formulae:

area of sector AOB = (1/2)r^2a
area of triangle AOB = (1/2)r^2sina

so area of segment made by AB is area of sector - area of triangle
segment = (1/2)r^2a - (1/2)r^2*sina
segment = (1/2)r^2(a-sina)

We need the angle a!

Draw a vertical line up from O to the mid-point of AB, call this point C. Now look at the right angled triangle AOC....

OA = radius = 1.105m
OC = 1.14-1.105m = 0.035m
angle AOC = b

so cosb=OC/OA
therefore b = 88.18489degrees

so angle a = 2b = 176.3697 degrees.

Now to convert this to radians....

360 degree = 2pi radians
so 1 degree is (2pi)/360 radians
hence 176.36 degrees is 3.07823 radians (to however many dp you want)

so,

segment above balls = (1/2)(1.105)^2(3.07823-sin(3.07823))
 = 1.8406m^2.

As cylinder area = pi*r^2
total Area = 3.83596m^2

then area of balls = 3.83596-1.8406 = 1.99546m^2

Hope this helps...Please check my working though, as i was rushing (i am at work), but the approach is correct.


jon.