# Calculus problem

## Homework Statement

A trough is 5 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of y=x^{2} from x=-1 to x=1 . The trough is full of water. Find the amount of work in foot-pounds required to empty the trough by pumping the water over the top. Note: The weight of water is 62 pounds per cubic foot.

## The Attempt at a Solution

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tiny-tim
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Welcome to PF!

Hi Tranquility13! Welcome to PF! Show us what you've tried, and where your'e stuck, and then we'll know how to help you! tiny-tim
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… one step at a time …

A trough is 5 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of y=x^{2} from x=-1 to x=1.
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)?

get that far, and then we'll sort out the integral! 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.

tiny-tim
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Hi greenandblue!

(are you the same as Tranquility13? … have a square-root anyway: √ )
dW=y*dF=y*62*5*2*√y*dy
Nooo … you have to pump the water over the top! That's what I thought I was doing... well my think was that if it took the force dF for the particular sliver dy, then it would take dF * the height of each sliver (distance = y, from 0, the top, to 1, the bottom), and do that for every sliver dy.

Since the water level is equal to the top of the trough then no further work would be required to get it out.

Of course I know I am off somewhere here, just not sure where.....

I'm new to this site, I didn't change my name from T.13, but I certainly appreciate your help!

tiny-tim
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I'm new to this site, I didn't change my name from T.13, but I certainly appreciate your help!
Hi greenandblue! erm … you're really not supposed to do other members' problems for them …

we'd better wait for Tranquility13 to catch up. we'd better wait for Tranquility13 to catch up. 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.

I was continuing her question b/c it is similar to one I have, which also deals with emptying a trough of the shape y=x^8 by 4ft long by 1ft tall.

The way I'd approach the problem is to first think about how you'd find the volume, and then remember that the work needed to pull out the water at each point is proportional to the height of the water at that point.

tiny-tim
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… oops!

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.
oops! You're absolutely right … my apologies! ok … the distance to the top of the trough isn't y, it' s 1 - y. ok … the distance to the top of the trough isn't y, it' s 1 - y. Ok, I got it!, I was considering that before but now it's clear and makes sense. Thanks for your help.