Answer to Question #228236 in Calculus for Mathpoint

Question

A hemispherical tank that is 12 ft in diameter is filled with a liquid that weighs 100 lb/ft^3. Find the work done in lowering the liquid by4 ft if it is expelled at a point 3 ft above the top of the tank.

Answer 1

Consider the sketch below

Here the radius of the hemisphere is $\frac{12}{2}ft = 6ft$

$\therefore 6^2=y^2+p^2\\ p^2=36-y^2$

The weight of a slab of liquid at the depth y of with Δy

$= \pi p^2Δy *100 U_0\\ = 100\pi (36-y^2)Δy U_0$

Work done to pump the upper 4 ft of the liquid to 3 ft above the tank

$W= \int_0^4 100 \pi (36-y^2)(y+3)dy\\ W= 100 \pi\int_0^4 (36y+108-y^3+3y^2)dy\\ W= 100 \pi[ (36y+108-y^3+3y^2)]_0^4\\ W= 59200 \pi ft.lb$