Q. 17 - Calculus

Question

Find the amount of labour necessary to completely drain the water from the hot tub's top, which has an $8$-foot diameter and a $3$-foot depth.

Step-by-Step Solution

Answer

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

Step 1: Given information

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

Step 2: Calculation

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 $V_k=\pi (4{)}^2\Delta y$, i.e., $V_k=16\pi \Delta y$ cubic feet and the water density is $\omega =62.4$ pounds per cubic foot.

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

Since $V_k=16\pi \Delta y,W=\omega Vd$ becomes $W=16\pi \omega d\Delta y.$

Substitute $d_k=3-y_k^{*}$ and $\omega =62.4$ in $W=16\pi \omega d\Delta y$ to obtain $W=16\pi (62.4)\left(3-y_k^{*}\right)\Delta y.$

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

$W=16\pi (62.4)\left(3-y_k^{*}\right)\Delta y.$

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

As $n\to \infty ,W=\sum _{k=1}^{n}16\pi (62.4)\left(3-y_k^{*}\right)\Delta y$ becomes a definite integral.

Step 3: Further Calculation

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 $y=3.$

$$\begin{gathered} W={\int }_0^{3}16\pi (62.4)(3-y)dy \\ =16\pi (62.4){\int }_0^3(3-y)dy \\ =16\pi (62.4)\left[3{\int }_0^{3}dy-{\int }_0^{3}ydy\right] \end{gathered}$$

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

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

$$\begin{gathered} W=16\pi (62.4)\left[3{\int }_0^{3}dy-{\int }_0^{3}ydy\right] \\ =16\pi (62.4){\left[3y-\frac{y^2}{2}\right]}_0^3 \\ =16\pi (62.4)\left[3\left(3\right)-\frac{(3{)}^2}{2}\right] \end{gathered}$$

$=14,114.55$

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