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

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    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.
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    Quote Originally Posted by twittytwitter View Post
    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<br />

    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
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