Q 52 - Calculus

Question

Find the work described in each of Exercises 49–60

The work required to pump all of the water out of the top of a spherical tank with a radius of 27 feet.

Step-by-Step Solution

Answer

The work required to pump all of the water is 138,908,319 foot pounds .

Step 1: Given Information

The work required to pump all of the water out of the top of a spherical tank with a radius of 27 feet.

Step 2: Diameter and radius

The radius of spherical tank is 27 feet. Diameter is 2 times radius 

Diameter = 2 times 27 = 54 feet

Consider the bottom of the tank is at y=54 feet and top of the tank is y=0

Thin slice of the tank is change in y that is delta y

Draw a horizontal representation y_k 

Apply Pythagorean theorem to find out radius in terms of y 

$$\begin{gathered} {27}^2={r_k}^2+{(27-y_k)}^2 729={r_k}^2+729-54y_k+{y_k}^2 \\ Subtract {r_k}^2 on both sides \\ -{r_k}^2+729=+729-54y_k+{y_k}^2 \\ Subtract 729 from both sides \\ -{r_k}^2=-54y_k+{y_k}^2 \\ Divide both sides by -1 \\ {r_k}^2=54y_k-{y_k}^2 \\ {r}_k=\sqrt{54y_k-{y_k}^2} \end{gathered}$$

Step 3: Volume

The slice is a disk that is in the form of circle 

The slice of the disk volume is ${\mathrm{\pi r}}^2∆\mathrm{y}$

The density of the water is 62.4 pounds 

The work required to lift and object is 

$$\begin{gathered} W= Fd=\omega Vd \\ W=62.4(\mathrm{\pi r}∆\mathrm{y})\mathrm{d} \end{gathered}$$

The work required to lift an object at distant d = y_k is 

$$\begin{gathered} W=62.4(\mathrm{\pi r}∆\mathrm{y})\mathrm{d} \\ \mathrm{W}=62.4({\mathrm{\pi r}}_{\mathrm{k}}∆\mathrm{y}){\mathrm{d}}_{\mathrm{k}} \\ {r}_k=\sqrt{54y_k-{y_k}^2} \\ W=62.4(\pi {\left(\sqrt{54y_k-{y_k}^2}\right)}^2 y_k∆\mathrm{y}) \\ W=62.4\mathrm{\pi }(54{\mathrm{y}}_{\mathrm{k}}-{{\mathrm{y}}_{\mathrm{k}}}^2){\mathrm{y}}_{\mathrm{k}}∆\mathrm{y} \end{gathered}$$

Work required to pump out   the slice of water is 

$W=62.4\mathrm{\pi }(54{\mathrm{y}}_{\mathrm{k}}-{{\mathrm{y}}_{\mathrm{k}}}^2){\mathrm{y}}_{\mathrm{k}}∆\mathrm{y}$

Step 4: Total work required

Work required to pump all the water completely , take integral from y=0 to 54

$$\begin{gathered} {\int }_0^{54}62.4\mathrm{\pi }(54\mathrm{y}-2^{}\mathrm{y}){\mathrm{y}}_{\mathrm{k}}\mathrm{dy} \\ 62.4\mathrm{\pi }{\int }_0^{54} (54\mathrm{y}-{\mathrm{y}}^2)\mathrm{ydy} \\ 62.4\mathrm{\pi }{\int }_0^{54} (54{\mathrm{y}}^2-{\mathrm{y}}^3)\mathrm{ydy} \\ 62.4\mathrm{\pi }{\left[54\frac{{\mathrm{y}}^3}{3}-\frac{{\mathrm{y}}^4}{4}\right]}_0^{54} \\ 62.4\mathrm{\pi }\left[54\frac{{\left(54\right)}^3}{3}-\frac{{\left(54\right)}^4}{4}\right] \\ 138,908,319 \end{gathered}$$

The work required to pump all of the water is 138,908,319 foot pounds