A water trough, 6 metres long, has uniform cross section in the shape of a semicircle with horizontal diameter 2 metres. It contains water to a depth of x metres. THe centre of the circle is at O and the triangle POQ is isosceles with angle POQ = @.

I've managed to get parts. But am stuck..

a. Find the area of the segment that is filled with water.

A = 0.5(@-sin@)

b. Find the volume of water in trough.

V = 6*A = 3(@-sin@)

c. If x metres is the depth of the water in the trough, find

i. x in terms of @

ii. dx/d@

iii. dx/dt in terms of @ and d@/dt

d. Find dV/dt.

dV/dt = 3(1-cos@)*(d@/dt)

e. If d@/dt = 0.05 rad/min, when @ = pie/3, find

i. Rate in m^3 per min at which volume of water is increasing.

ii. Rate in metres per min at which water level is rising.