just found a really helpful sample problem online. thanks anyway.
might be posting another problem here in a few
ok, i drew out a diagram and I have no idea how or where to start this problem. Please help guide me so I can learn how to do it...
A water tank has the shape of an inverted circular cone with a diameter of 4 feet and a height of 6 feet. If the tank is full, find the work required to empty the tank through a pipe that extends 3 feet above the top of the tank. Use 62.5 as the weight-density of water. Give your answer in foot-pounds.
W = weight-density in lbs per cubic foot
A = cross-sectional area of a representative horizontal "slice" of liquid in terms of y
L = lift distance of the representative "slice" in terms of y
T = "slice" thickness , dy.
let the vertex of the cone be at the origin ... one side of the cone is modeled by the linear equation ...
"slice" radius = , so the cross-sectional area of a representative "slice" is ,
the liquid resides between and ... your limits of integration.
set up the integral and find the work required.
thanks, ill be sure to remember WALT.
This is how I solved it, not 100% if its right but it seems it might be...
Weight of slab = density * volume
weight of slab = 500 pi dx
Distance = x + 3
Work = (x + 3)500pi dx
W = integral from 0 to 6 [(x+3)(500pi)]dx
W = 56,548.7 ft lbs