## anonymous 4 years ago A cylindrical tank of radius 2 m and height 4 m is full of water. Find the work required to pump all of the water to a point that is 2 m above the top of the tank. The density of water is = 1000 kg/m3 and the acceleration of gravity is g = 9:8 m/s Define clearly all the variables that enter into your solution and provide a drawing which shows their meaning

1. saifoo.khan

What part u dont understand?

2. anonymous

ok ... we have to make small cylinders with thickness dx. so the volume will be 4pi dx and to find the work you find the mass and the force and then to find the work you multiply it by the hieght and integrate it... now what are your limits of integration ?

3. amistre64

4. amistre64

|dw:1328655961951:dw|

5. amistre64

the distance changes as we integrate; the weight of the water is a factor of volume ....

6. amistre64

and i forgot the 2m above so that should be included in overall distance as well

7. anonymous

oh.. ok , now I remembered , from the bottom to the top of the cylinder is 4m and we want to take it 2m above so we integrate from 0 to 6

8. amistre64

http://www.youtube.com/watch?v=oWcN96kcbkM&feature=related yep, this might help, patrick is awesome

9. amistre64

cable rope pull is same concept as water pumped

10. amistre64
11. amistre64

volume of water = pi * (2m)^2 * (dy)m * (1000) = 4000pi(distance moved)dy distance moved is: 6-y $\int_{0}^{4}4000pi(6-y)dy$sounds right to me question is, is it right?

12. amistre64

im not sure what the gravity is for; unless this is spose to be measured in newtons

13. anonymous

yup, that's what I got , the gravity is for the force W=F.d F=mg m=(roh)V but the only thing I don't get is that when we bring the water 2m above the cylinder then why don't we integrate from 0 to 6 instead of 0 to 4 ?

14. amistre64

because the weight of the water is not fluctuatung in that last little bit i assume. or rather, the amount of water that is above 4 is zero, so that extra 2 feet only affects the distance to move the water