Integrals To Find Amount of Work Needed
I can't figure this out, I know how to do integrals, My problem is I can never set up the equations correctly. Can someone please help me solve this.
Toxic waste (a homogeneous liquid whose density is 300 kg/m^3) is stored in three buried containers. Each container, is 10 meters tall. The top of each container is at ground level. Each container has the same volume. One container is a circular right cone, the next is an inverted circular right cone, and the last a cylinder with radius 2m. Which container needs the least amount of work to empty? Which needs the most? Justify by calculating the work and then explaining why your answers make sense.
Re: Integrals To Find Amount of Work Needed
EDIT: And after writing all the stuff below, I realized you asked this question in 2006... Maybe it will help someone else...
These kind of problems which involve finding the work to empty tanks all have the same steps to solve.
1) Introduce a vertical x-axis (usually making it positive going down simplifies the problem)
2) Cut the relevant interval (in this case [0,10]) into n sub-intervals with size
3) Find the radius of the solid as you move down it, in terms of x. For the cylinder it is always 2m, but for the cones the radius changes as you go down. You'll usually have to use similar triangle ratios to do this.
4) Use the radius to find an approximate volume for any given sub-interval with size in terms of x. You can do this by assuming the shape of the solid in the sub-interval is approximately a cylinder since the change in x is very small. So Note h in this case is just
5) Using the equation , where is the density, find the mass of the toxic waste in this sub-interval.
6) Find the force needed to move this amount of toxic waste in the sub-interval to the top of the solid by using, F=mg, where g is the acceleration of gravity. You can use this since gravity is the only force opposing the emptying of the solid.
7) Find the work,W, done in moving the toxic waste in this sub-interval by W=Fd, where d is the distance you had to move the waste, (usually d=x, since you have to move the waste in any sub-interval a distance of x. If you sub in all the previous equations you get
8) In 7 you found the approximate work of moving the waste in a sub-interval to the top of the solid. To find the total work, you have to add up all of the work for all n sub-interval. You get Riemann sum, and letting n go to infinity will give you the integral. In this case it would be:
Try to follow the above steps with one of the cone solids, since the cylinder is easier.
As for which one requires the least work. You can figure that out by logic. For the cone, the mass of the waste in it is mostly at the bottom, so it would need the most work to empty, since you will be moving the waste a longer distance. For the inverted cone, the mass is at the top, so it needs the least work to empty. The cylinder has equal mass throughout, so the work required for it is in between the two cone shaped solids. So in order of increasing work: inverted cone, cylinder, cone.