So given the question "A conical tank of radius $5\mathrm{m}$ and height $10\mathrm{m}$ is filled with a liquid whose density is $3.42 \mathrm{kg/m^3}$. how much work is needed to lift the liquid out of the tank?" The answer can be found by taking the integral $$\int_{0}^{10} 3.42 \cdot \pi \cdot 9.8 \cdot (10-h) \cdot (h/2)^2\ dh $$ However, I'm confused about one thing. The equation $$ (10-h)\cdot (h/2)^2\cdot 9.8\cdot 3.42\cdot \pi \cdot dh $$ gives us the work required to lift a disc of height $dh$ out of the tank. Why does the integral of this expression within the limits 0-10 give us the work required to lift all the water out of the tank?
Thanks, newtondiedavirgin
a) the tank is a cone with the vertex at the top, and base on ground;
b) $h$ is actually the distance from the vertex, the height being $H=10-h$, at which the radius is $h/2$;
c) the integrand is the potential energy acquired by disk $\rho \pi r^2 dH$ at height $H$;
d) the integral is the work needed to fill the tank (= potential energy), i.e. for the liquid to start and pour from the top, with the substitution $h$ for $H$.