Calculus II Question?

Question

Have a homework question that I'm having touble solving even question so no correct answer in book. Not to turn in. Class grade is from quiz & exams only.

Water tank shape of inverted circular cone. radius = 5 ft, height = 20 ft, filled to 16 ft with water at a weight of 62.5 lbs/ft3. How much energy to empty tank by forcing water over top of tank.

62.5(pi/16)(20-X)^2 (deltaX)

32ft/s^2(62.5)(pi/16)((20-X)^2... (deltaX)

(125pi)(20-X)^2 (deltaX)

125pi(integral from 4 to 20) 400x-40x^2+x^3

125pi(200X^2-(40X^3/3)+(X^4/4) | 20 to 4

125pi(32768/3) = 4.289 x 10^6 ft lbs

I'm pretty sure this is wrong because example in book is much larger metric tank (r=4m, h=10m, water = 8m, 3.4 x 10^6 J) with a smaller Joule energy requirement. Mabye no one to help but any input is appreciated.

Answer 1

First you have water density 62.5 lb/ft^3 not weight of the water.
Second do not worry about the answer in the book. First is the soundness of the idea and the governing equation

The idea is to remove differential volume over the top of an inverted cone and the sum it up as the water leaves over the top.

The equation is

W= integral((H-h) p pi r^2 dh) from h=H to h=0

W= work done
H- height of the cone (in your case it 10 feet)
h- height variable
r - radius of the cone (there is a relationship between r and h and to be used during integration)
p - water density
dh - differential thickness of the disk of water

Since r and h are related as a line y=mx
we have h=mr m=H/R=20/5=4
so h=4r or r =h/4

We have

W=(p pi )integral((H-h) (h/4)^2 dh) from h=20 to h=0

W=(p pi )[(Hh^3)/12 - (h^4)/16] evaluated from 20 to from 20 to 0

Just do the numbers and conversions if necessary.

Answer 2

Looks to me like you might have your coordinates mixed. Note that at x=0 you have a diameter of 20 but you're doing the integral from 4 to 20.

Hmm, by 'inverted cone': where is the vertex? You can probably change the variable of integration by letting y=20-x . It will make the intergral easier.

(125pi)(20-X)^2 (deltaX)
this becomes 125[400 -40x+x^2]
the indef integral becomes
125[400x -20x^2 + x^3/3] this differs from yours.

Hope that helps.