Answer
$\approx$ $1.04\times10^5$ $ft-lb$
Work Step by Step
Let $x$ measure depth (in feet) below the spout at the top of the tank.
A horizontal disk-shaped of water $Δx$ ft thick and lying at coordinate x has radius $\frac{3}{8}(16-x)$ ft and volume ${\pi}r^{2}Δx$ = ${\pi}\frac{9}{64}(16-x)^2Δx$ $ft^{3}$.
It weighs about $(62.5)\frac{9\pi}{64}(16-x)^2Δx$ $lb$ and must be lifted $x$ ft by the pump, so the work needed to pump it out is about
$(62.5x)\frac{9\pi}{64}(16-x)^2Δx$ $ft-lb$
The total work required is
$W$ = $\int_0^8{(62.5x)\frac{9\pi}{64}(16-x)^2}dx$
= $(62.5)\frac{9\pi}{64}\int_0^8{(256x-32x^2+x^3)}dx$
= $(62.5)\frac{9\pi}{64}{(128x^2-\frac{32}{3}x^3+\frac{1}{4}x^4)}|_0^8$
= $(62.5)\frac{9\pi}{64}(\frac{11264}{3})$ $\approx$ $1.04\times10^5$ $ft-lb$
