- #1
Punkyc7
- 420
- 0
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(sqrt(4-y^2))^2 from 0-1
D=2-x
so W=rho*V int 2-x from 0-1
i get 11pi rho g/3 and the answer is suppose to be 9 rho pi/4
im not sure where i am going wrong but i thing it has to do with my integration
W=FD
V=integral of surface area * height
F= rho *V
V is the integral of pi(sqrt(4-y^2))^2 from 0-1
D=2-x
so W=rho*V int 2-x from 0-1
i get 11pi rho g/3 and the answer is suppose to be 9 rho pi/4
im not sure where i am going wrong but i thing it has to do with my integration