Q. 17 - Calculus

Question

Set up and solve definite integrals to answer each of the following questions. 

Find the work required to pump all of the water out of the top of a cylindrical hot tub that is $8$ feet in diameter and $3$ feet deep.

Step-by-Step Solution

Answer

$14,114.55 pounds$.

Step 1. Given Information.

Find the work required to pump all of the water out of the top of a cylindrical hot tub that is 8 feet in diameter and 3 feet deep.

Step 2. Diagram.

Use the fact that the radius of a circle is half of its diameter to obtain the radius of the cylinder. The radius of the cylinder is $\frac{1}{2}·8 ft=4ft$.

Consider that the top of the tank is at height $y=3$ and the bottom of the tank is at height $y=0$. Draw a diagram that shows a thin representative slice of the tank at some point $y_k^{*}$ from the bottom.

Step 3. Formulation.

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

The slice is a disk with a volume,

 $$\begin{gathered} V_k=\mathrm{\pi }{\left(4\right)}^2∆\mathrm{y} \\ {\mathrm{V}}_{\mathrm{k}}=16\mathrm{\pi }∆\mathrm{y} \end{gathered}$$ cubic feet

and the water density is $\omega =62.4$pounds per cubic foot.

The work required to lift an object of weight $F$ through a distance $d$ is given by

$W=Fd=\omega Vd$.

Therefore, the work required to pump out the representation slice of water is

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

Step 4. Calculation.

The integral is defined as,

$$\begin{gathered} W=16\pi \left(62.4\right)\left(3-y_k^{*}\right)∆y \\ W= 16\pi \left(62.4\right){\int }_0^3\left[3-y\right]dy \\ W= 16\pi \left(62.4\right){\left[3y-\frac{y^2}{2}\right]}_0^3 \\ W=14,114.55 \end{gathered}$$