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 :)

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= ?

But, if the tank is full when the work begins, then the distance is only a distance (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?

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

You're right



Do that and you're finished.

thanks alot for ur help