did you learn anything from this post?
http://www.mathhelpforum.com/math-he...k-pumping.html
A circular swimming pool has a diameter of 16 m, the sides are 4 m high, and the depth of the water is 3.5 m. How much work (in Joules) is required to pump all of the water over the side? (The acceleration due to gravity is 9.8 and the density of water is 1000 .)
so i tried
volume=64pi dx
weight= vol *density
64000pidx
integral .5 to 4 64000pidx
did you learn anything from this post?
http://www.mathhelpforum.com/math-he...k-pumping.html
Won't the work required to be done just be the gain in in gravitational PE of the entire water?
The total mass of the water= Volume of the water(density)
= 1000(pi)(64)(3.5)=m(say.)
Now when the water is pumped over the side, the gain in gravitational PE is mg(4-3.5) = mg(0.5)
=3448220.16 J ??????
Since the density is constant and the cross sections are constant
we don't need an integral.
So we think of the mass as being concentrated at the center of gravity
here y= 1.75 and the distance 4 - 1.75 = 2.25 m
weight is pi * 64*1000*9.8*3.5
Calculate weight * dist = 64*1000*9.8*3.5*pi*2.25
If you do integrate the limits are 0 to 3.5
you have 64*1000*9.8*pi integral[(4-y)dy]