# Application of integration - Work

• Jul 22nd 2012, 08:52 AM
hayjude99
Application of integration - Work
I was wondering if I could elicit some help with this problem. I feel as if I understand it but cannot get the right answer.

http://i.imgur.com/6YdBs.png

It is number 21.
These are the steps I took:
• ΔF=(62.4ft/cm^3)(volume)
• =(62.4)(1/3 π r2h)
• Since this is a cone, the radius isn't constant. I imagined that the come was made by revolving the integral x=2/3 y about the y axis, which then gives the r at any given height, Δy
• ΔF=62.4(1/3 π (2/3y)2Δy)
• = 9.244π* y2Δy
• Then
• ΔW=ΔF*d
• Let d= 6-y
• ΔW=(9.244π* y2Δy)(6-y)
• W=9.244π*int [0, 6] 6y2-y3
• =9.244π(432-324)
• =998.352π

However the correct answer is 2995.2π...

By the way I'm doing this on my phone so if the formatting is terrible, I'm sorry I'll try and revise it.
• Jul 22nd 2012, 09:25 AM
skeeter
Re: Application of integration - Work
work = the integral of WALT

W = weight density of the liquid ... $\displaystyle 62.4 \, lbs/ft^3$

A = cross-sectional area of a representative horizontal "slice" of liquid as a function of y ... $\displaystyle \pi \left(\frac{2y}{3}\right)^2$

L = lift distance of a that representative slice as a function of y ... $\displaystyle (6 - y)$

T = slice thickness ... $\displaystyle dy$

the limits of integration are determined by where the horizontal slices reside w/r to y

work in ft-lbs = $\displaystyle \int_0^6 62.4\pi \left(\frac{2y}{3}\right)^2(6-y) \, dy$
• Jul 22nd 2012, 09:44 AM
hayjude99
Re: Application of integration - Work
Thanks, I had noticed I needed a factor of 3 to get to the right answer. My book doesn't seem to use the WALT method, do you know how this ties in to using the W=Fd method? I just don't understand why both methods don't work.

Thanks again
• Jul 22nd 2012, 10:07 AM
skeeter
Re: Application of integration - Work
Quote:

Originally Posted by hayjude99
Thanks, I had noticed I needed a factor of 3 to get to the right answer. My book doesn't seem to use the WALT method, do you know how this ties in to using the W=Fd method? I just don't understand why both methods don't work.

both methods are the same ...

the amount of work required to lift a single horizontal slice of water = (weight of that slice)(vertical displacement of that to the top of the cone)

$\displaystyle dW = \color{red}{F} \cdot \color{blue}{\Delta y}$

$\displaystyle dW = \color{red}{mg} \cdot \color{blue}{\Delta y}$

$\displaystyle dW = \color{red}{\rho \cdot \pi \left(\frac{2y}{3}\right)^2 \, dy \cdot g} \cdot \color{blue}{(6-y)}$

note that $\displaystyle \rho \cdot g$ = weight density of water

and $\displaystyle \pi \left(\frac{2y}{3}\right)^2 \, dy$ is the volume of a representative slice
• Jul 22nd 2012, 11:25 AM
hayjude99
Re: Application of integration - Work
Oh, I see what happened. I went into the problem think that the volume of the cone was 1/3 π r2h and then attempted to find the r in terms of y. But that radius in terms of y, when integrated , actually gives me the volume.
Thank you