Answer to Question #198258 in Calculus for Ben

Question

Work

Problem 5: You are in charge of the evacuation and repair of the storage tank shown in the figure below. The tank is a hemisphere of radius 10 ft and is full of benzene weighing 56 lb / ft^3 . A firm you contacted says it can empty the tank for 1/2 ¢ per foot-pound of work. Find the work required to empty the tank by pumping the benzene to an outlet 2 ft above the top of the tank. If you have $5,000 budgeted for the job, can you afford to hire the firm? 

Answer 1

Solution :-

The typical slab between the planes at $y \ and \ y+\Delta y$  has a volume of about

$\Delta v=\pi({radius)}^2$

$=\pi{(\sqrt{100-y^2})}\Delta y \\ = \pi{(\sqrt{100-y^2})}\Delta yft^3.$

The force is  $f(y)=\frac{56lb}{ft^3}.\Delta v$

$=56\pi(100-y^2)\Delta lb$

The distance through which f(y) must act to lift the slab to the level of 2ft above the top of the tank is about (12-y)ft.

So the work done is $\Delta w \approx56\pi(100-y^2)(12-y)\Delta lb-ft$

The work done lifting all the slabs from y = 0 ft to y = 10ft is approximately.

$w=\sum_0^{10} 56\pi(100-y^2)(12-y)\Delta ylb.ft$

Taking the limit of these Riemann sums we get

$w=\int_0^{10}56\pi(100-y^2)(12-y)dy$

$=56\pi[1200y-\frac{100y^2}{2}-\frac{12y^3}{3}+\frac{y^4}{4}]_0^{10}$

$=967.6111ft-lb$

the would cost $(0.5)(967611)=483,805 ¢ =\$4838.05$