91Ó°ÊÓ

Hot tub's top has $8$ feet in diameter and$3$feet depth.

To determine the radius of the cylinder, use the rule that a circle's radius is equal to half of its diameter.

The cylinder's radius is $\frac{1}{2}.8ft=4ft$.

Assume that the tank's top is at height $y=3$and its bottom is at height $y=0.$

Create a diagram that displays a narrow slice of the tank that is representative and is at the same height $y_k^{*}$from the bottom.

The slice at $y_k^{*}$needs to be moved upwards by $d_k=3-y_k^{*}$should be pushed out of the tank in units.

This slice is a disk with volume $V_k=\mathrm{Ï€}(4{)}^2\mathrm{Δ}y$, i.e., $V_k=16\mathrm{Ï€}\mathrm{Δ}y$cubic feet and the water density is $\mathrm{Ó\neg }=62.4$pounds per cubic foot.

The formula for the amount of work needed to lift an object of weight F across a distance d is $W=Fd=\mathrm{Ó\neg }Vd.$

Since $V_k=16\mathrm{Ï€}\mathrm{Δ}y,W=\mathrm{Ó\neg }Vd$becomes $W=16\mathrm{Ï€}\mathrm{Ó\neg }d\mathrm{Δ}y.$

Substitute $d_k=3-y_k^{*}$and $\mathrm{Ó\neg }=62.4$in $W=16\mathrm{Ï€}\mathrm{Ó\neg }d\mathrm{Δ}y$to obtain $W=16\mathrm{Ï€}(62.4)\left(3-y_k^{*}\right)\mathrm{Δ}y.$

Consequently, the effort needed to pump out the representative water sample is

$W=16\mathrm{π}(62.4)\left(3-y_k^{*}\right)\mathrm{Δ}y.$

The amount of labour needed to remove all the water from the spherical tank's top is about $W=\underset{k=1}{\overset{n}{∑}}16\mathrm{π}(62.4)\left(3-y_k^{*}\right)\mathrm{Δ}y.$

As $n→∞,W=\underset{k=1}{\overset{n}{∑}}16\mathrm{π}(62.4)\left(3-y_k^{*}\right)\mathrm{Δ}y$becomes a definite integral.

To calculate the amount of work needed to pump all the water out of the tank's top, add up the slices from $y=0$ to $y=3.$

The power rule for differentiation states that $∫x^{n}dx=\frac{x^{n+1}}{n+1},$ where $n$ is a real number.

Use power rule to evaluate the integral $16\mathrm{π}(62.4)\left[3{∫}_0^{3}dy-{∫}_0^{3}ydy\right].$

$=14,114.55$

The required work is$14,114.55$foot pounds.