Find the integral for pumping water out of a cone-shaped tank.

[shamieh]

All I need to do for this problem is set up the integral...Can someone tell me how to do that?

A tank has the shape of an inverted circular cone with height 10m and base with radius 1m. The tank is filled with water to a height of 8 m . Find the work required to empty the tank by pumping all of the water over the top.NOTE: I just need to set up the integral, I don't actually have to calculate the problem.

 

[MarkFL]

Here are two threads that deal with this kind of problem:

http://mathhelpboards.com/questions-other-sites-52/kendra-ns-question-yahoo-answers-regarding-work-done-empty-conical-tank-4661.html

http://mathhelpboards.com/questions-other-sites-52/domenics-question-yahoo-answers-regarding-computing-work-empty-tank-8541.html

 

[HallsofIvy]

All I need to do for this problem is set up the integral...Can someone tell me how to do that?

A tank has the shape of an inverted circular cone with height 10m and base with radius 1m. The tank is filled with water to a height of 8 m . Find the work required to empty the tank by pumping all of the water over the top.NOTE: I just need to set up the integral, I don't actually have to calculate the problem.

From the side the cone looks like a triangle and we can use similar triangles. With height h, the distance from the center of the cone to the side, r, we have r/h= 1/10 so that r= h/10. The area of the disk at that height is $$\pi r^2= \pi h^2/100$$ and the volume of a thin 'layer of water', with thickness dh is $$\frac{\pi}{100}h^2 dh$$. Taking $$\delta$$ to be the density of water, it's weight is $$\frac{\pi\delta}{100}h^2dh$$. Lifting that from height h to height 10m requires $$\frac{\pi\delta}{100}h^2(10- h)dh$$ Joules of work. Integrate that from h= 0 to h= 8.