Question 1178402
Refer to the drawing of the cross-section of the pipe. The radius, R of the
pipe is 0.9/2. The size of the angle BAC is A = 2*acos(0.15/R), and the area of
the sector subtended by this angle is As = A*R^2/2
If we subtract the area of the triangle ABC from As, this will give us the area
of the empty space
The cross sectional area of the liquid in the pipe will be the total cross 
sectional area minus the area of empty space
The area of triangle ABC = At = 0.15*sqrt(0.45^2 - 0.15^2)
Putting in all the numbers we have As = 2.46191*0.45^2/2 = 0.2493 and
At = 0.0636
Thus the area of liquid is pi*0.45^2 - (0.2493 - 0.0636) -> 0.4505
Finally the volume is the area times the length:
V = 0.4506*3 = 1.3516 m^3 = 358.18 gallons
*[illustration pipe.png]