Work: How much work to pump water six feet above the top?

[shivers20]

A resrvoir shaped like a right circular cone, point down, 20ft across the top and 8ft deep, is full of water. How much work does it take to pump the water 6ft above the top?

. . .W = fd
. . .d = x + 6
. . .V = pi r^2 dx
. . .r/8 - x = 10/8
. . .r = 10/8 (8 - x)

. . .W = }0 pi (10/8) (8-x)^2 (x+6) dx
. . . . .. .{8

. . .pi (10/8)^2 (8 - x^2) (x + 6)dx

Is this the final answer? Or do I need to go further? If so, how?

 

[tkhunny]

Wow! You'll have to try that again.

1) How did you get from (10/8)(8-x)^2 to [(10/8)^2]*(8-x^2)? There's some seriously mystifying algebra and notation going on in there. I think you had the right idea, [(10/8)(8-x)]^2, but that is not what you wrote and it certainly isn't where you ended up.

2) Where is the weight of the water? You can't know the Force without that. Otherwise, it takes the same work to move alcohol or to move mercury. That can't be right.

3) You will have to evaluate the integral before you think you are done.

Slowly...deliberately...carefully...

 

[skeeter]

let the origin be the bottom of the conical tank. the slanted side of the tank is defined by the line y = (4/5)x

for these "pumping" type problems, work = Integral of WALT, where
W = weight density of the fluid
A = cross-sectional area of a representative "slice" of liquid
L = amount of lift for that representative slice
T = slice thickness

for water, W = 62 lbs/ft³
A = pi*x^2 ... in terms of y, A = (25pi/16)y²
L = 14 - y
T = dy

\(\displaystyle \L work = \int_0^8 62*\frac{25\pi}{16} y^2 (14-y) dy\)

evaluate the definite integral to determine the work done in ft-lbs.