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 r
^{2}h) - 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)
^{2}Δy) - = 9.244π* y
^{2}Δy - Then
- ΔW=ΔF*d
- Let d= 6-y
- ΔW=(9.244π* y
^{2}Δy)(6-y) - W=9.244π*int [0, 6] 6y
^{2}-y^{3} - =9.244π(432-324)
- =998.352π

However the correct answer is 2995.2π...

Thank you in advance!

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.

Re: Application of integration - Work

work = the integral of **WALT**

W = weight density of the liquid ...

A = cross-sectional area of a representative horizontal "slice" of liquid as a function of y ...

L = lift distance of a that representative slice as a function of y ...

T = slice thickness ...

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

work in ft-lbs =

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

Re: Application of integration - Work

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 r^{2}h 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