Water pump problem

We can write radius of a sliced cross section of the frustum as $r+\dfrac{(R-r)*x}{h}$ as a function of x, Hence

Area of circles formed as a function of x is = $\pi*(r+\dfrac{(R-r)*x}{h})^{2}$

Now integrating area of circles of radius given by this function with bounds h and 0 gives the volume . $\displaystyle\int_0^h \pi*(r+\dfrac{(R-r)*x}{h})^{2} dx$ , substituting r=6 ,R=12 and h=17, and multiplying with weight density we get force acting on each differential volume of water. $\ \displaystyle\int_0^h 62.5*\pi*(6+\dfrac{6*x}{17})^{2} dx$

Now Consider a differential volume of the frustum at x, this volume must be lifted (17-x) distance. We have Force = $\ \displaystyle\int_0^h 62.5*\pi*(6+\dfrac{6*x}{17})^{2} dx$ and displacement equals (17-x),

W= $\ \displaystyle\int_0^h (17-x)*62.5*\pi*(6+\dfrac{6*x}{17})^{2} dx$ definite integral from 0 to h=17 gives 1872590 lb-ft.