The problem involves calculating the work required to empty a water trough that is 5 feet long and 1 foot high, with a vertical cross-section shaped like the graph of y=x² from x=-1 to x=1. The weight of the water is given as 62 pounds per cubic foot.
Some participants have offered guidance on how to approach the problem, including the need to consider the distance to the top of the trough. There is ongoing exploration of the correct setup for the integral needed to find the work.
There is a note that one participant is working on a similar problem involving a different shape for the trough, indicating a broader context for the discussion. Additionally, there are indications that the original poster has not engaged with the thread for an extended period.
Tranquility13 said:
tiny-tim said:Hi Tranquility13!
ok … divide the trough into horizontal slices, a distance y above the bottom, and with height dy.
what is the volume of this slice?
what is the weight of the water in it?
how much work is needed to lift that water to the top of the trough (ie to y = 1)?
Hi TinyTim (or someone else),
Maybe you can help me finish this problem:
A=l*w=5*2*sqrt(y)
dV=A*dy=5*2*sqrt(y)*dy
dF=p*dV=62*5*2*sqrt(y)*dy
dW=y*dF=y*62*5*2*sqrt(y)*dy
so then the integral is:
int{6200*y*sqrt(y)*dy, from y=0 to y=1}
But this fails to give me the right answer. Hopefully you can shed some light.
Thanks.
)greenandblue said:
greenandblue said:
I completely agree except in this case: T.13 has left her problem untouched since her last post on May 1st, 3.5 months ago. It seems that by now she has either solved the problem or lost interest in it.tiny-tim said:we'd better wait for Tranquility13 to catch up.
greenandblue said:
Ok, I got it!, I was considering that before but now it's clear and makes sense. Thanks for your help.tiny-tim said:ok … the distance to the top of the trough isn't y, it' s 1 - y.