Find the work in pumping the water out fo the tank

[Punkyc7]

a bowl shaped tank is in the shaoe of a hemisphere with a radius of 2 in. If the bowl is filled with water density rho to a depth of 1 in, find the work in pumping the water out fo the tank



W=FD

V=integral of surface area * height

F= rho *V
V is the integral of pi(r-x)^2 from 0-1
D=2-x

so W=rho*V int 2-x from 0-1

i get 7pi rho g/2 and the answer is suppose to be 9 rho pi/4

im not sure where i am going rong but i thing it has to do with my integration

we haven't done parametrics yet

 

[rl.bhat]

dV = π*r^2*dh.

r^2 = R^2 - h^2.

Hence V = π*Int[R^2 - h^2]*dh from h = 1 to h = 2.

Now proceed.

I think the answer is wrong.

 

[Punkyc7]

Why are your limits from 1-2 its only filled half way shouldn't it be 0-1 if your point of reference is from the bottm

 

[rl.bhat]

Why are your limits from 1-2 its only filled half way shouldn't it be 0-1 if your point of reference is from the bottm

While emptying the bowl, water level changes from 1 to 2, where h is measured from the center. When you write down the relation between R and h, h is measured from the center top surface of the bowl.

 

[Punkyc7]

ah ok i was doing it from the bottom so the distance is 1 + x now right