Need Help URGENT!!!

[mesabat]

I am coming across a problem in my math homework that completely defeats me every time I try to solve it. The problem is as follows:
Find the amount of work required to pump water out of a full horizontal cylindrical tank through a hole at the top. Use L for the length and R for the radius.
Any help is apreciated.

 

[skeeter]

general integral to calculate work required to vertically pump liquid upward from a tank (cylindrical, spherical, conical, whatever)

[MATH]W = \int_c^d D_w \cdot A(y) \cdot L(y) \, dy [/math]
where ...

[MATH]D_w[/MATH] is the weight density of the liquid

[MATH]A(y)[/MATH] is the cross-sectional area of a representative horizontal “slice” of liquid as a function of y

[MATH]L(y)[/MATH] is the lift distance of that representative slice as a function of y

[MATH]dy[/MATH] is the slice thickness

 

[Steven G]

if you show us your work then we can help you figure out what is going wrong with your work.

Please understand that this is a free math help forum and not a homework service site. That is to receive help we need to see your work so we know what type of help you need.

For the record if you followed the posting guideline for this forum you would have received help by now.

 

[HallsofIvy]

The way to think about this problem is to imagine many thin layers of water, disks of radius R so area \(\displaystyle \pi R^2\) and thickness "dx".
That will have volume \(\displaystyle \pi R^2dx\) so weight \(\displaystyle \pi\rho R^2dx\) where \(\displaystyle \rho\) is the density of water. The layer of water at height "x" above the bottom of the tank has to be lifted a distance h-x to the top of the tank. The energy necessary to do that is "force times distance" or \(\displaystyle \pi\rho R^2 (h- x)dx\). The energy necessary to lift all the water out of the tank is the "sum"of the energy necessary to lift all those layers. Of course, since this is a continuous block of water, not individual layers, that "sum" is really an integral, \(\displaystyle \int_0^h \pi\rho R^2 (h- x)dx= \pi\rho R^2\int_0^h (h- x)dx\).