# work needed to pump water out a tank

• Jan 31st 2009, 07:34 PM
alex83
work needed to pump water out a tank
a tank is formed by revolving the curve y=x^2 ; 0<=x<=4 ; around y-axis
if filled with water with weight 10000 N/m^3
find work needed to empty the tank by pumping it to the top ?

the problem is that i dont have the height to set the integral
is it right if i use the radius as 4?

this way ill get V=PI r^2 dy
F=10000*V
W=(?) *F
integral limits ==??

any help or hints pls :)
• Jan 31st 2009, 07:50 PM
Aryth
Here's a hint for the height:

If the farthest on the x-axis (horizontally) you can go is 4, then would it not be correct that the farthest on the y-axis (vertically) you can go is y= $x^2 = 4^2 = 16$ ?

But, if the tank is full when the work begins, then the distance is only a distance $x^2$ (starting at x=4) from the maximum height. So what do you think the height is?

And as a result, what does your work integral look like?
• Jan 31st 2009, 08:02 PM
alex83
actually my work integral is : Force*volume*hight of slab*dy ]with integral limit to cover the whole hieght

from ur hint i got the limit of integral from 0-16 (cuz im pumping out from the top)

*do u think its correct if i use it as follows

w= (16*PI)(10000)(16-y)dy [integral 0-16]???

thank you
• Jan 31st 2009, 08:09 PM
Aryth
You're right

$W = (16*10000)\pi \int_{0}^{16} [16 - y] ~dy$

Do that and you're finished.
• Jan 31st 2009, 08:12 PM
alex83
thank u
thanks alot for ur help
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