# Work it, I need a glass of water....Solving for F(h)

• Feb 2nd 2011, 07:44 AM
Work it, I need a glass of water....Solving for F(h)
Water is pumped into a spherical tank of radius 5 ft from a source located 2 ft below a hole at the bottom. Density of water=62.4. Find the work F(h) to fill the tank to a height h from the bottom of the tank.

I know F(0)=0 and F(10)=228708 by computing the work to fill the tank. It is clearly going to be a continuous function and it fills faster in the middle, the domain will be 0<=h<=10, but how do I find the equation for this? I know I can still use the same geometry I did for finding F(10), with center at the center of the sphere, but I am confused how to get a function out of this.
• Feb 2nd 2011, 09:33 AM
skeeter
Quote:

Water is pumped into a spherical tank of radius 5 ft from a source located 2 ft below a hole at the bottom. Density of water=62.4. Find the work F(h) to fill the tank to a height h from the bottom of the tank.

I know F(0)=0 and F(10)=228708 by computing the work to fill the tank. It is clearly going to be a continuous function and it fills faster in the middle, the domain will be 0<=h<=10, but how do I find the equation for this? I know I can still use the same geometry I did for finding F(10), with center at the center of the sphere, but I am confused how to get a function out of this.

consider a representative horizontal slice of water in the tank between the bottom and height $h$ ...

$dV = \pi \cdot x^2 dy$

let the side view of the sphere be the circle $x^2 + y^2 = 25$ ... $x^2 = 25-y^2$

$dV = \pi(25-y^2)dy
$

weight of the slice is $62.4 \cdot dV$

note that $h = y + 5$ , so the distance the slice is to be lifted is $(y+7)$

work to lift all the slices ...

$\displaystyle W(h) = 62.4 \int_{-5}^{h-5} \pi(25-y^2)(y+7) \, dy$