Q. 2 - Calculus

Question

Now suppose we want to pump all of the water out of the top of a 4-foot-high cylindrical tank with radius of 10 feet. It takes more work to pump out the water from the bottom of the tank than it does to pump it out from the top of the tank. With the given figure as a guide, use horizontal slices to set up and solve a definite integral that represents the work required to pump all of the water out of the top of the tank. 

Step-by-Step Solution

Answer

The work done to pump out the water from the tank is $478694$ meter kg.

Step 1. Given Information

The radius of the cylindrical tank is $10$feet

The height of the tank is $4$ feet.

The daigram:

Step 2. Find the weight of the water in the slice.

In the given diagram we can see that there is a horizontal slice has been cut whose radius is

 $10$feet or $3.048$ meter.

The thickness of the slice is $∆y$

So the area of the slice is:

$A=\mathrm{\pi }{(3.048)}^2$ square meter

And volume of the slice is $v=\mathrm{\pi }{(3.048)}^2∆\mathrm{x}$

Since density of the water is $1000$ kg per cubic meter.

Hence the mass of the water in the slice is;

$m=1000\mathrm{\pi }{(3.048)}^2∆\mathrm{y}$

The weight of the water in the slice is   $mg$ where $g=9.8m/s^2$

$$\begin{gathered} w=9.8\times 1000\mathrm{\pi }{(3.048)}^2∆\mathrm{y} \\ =91045\mathrm{\pi }∆\mathrm{y} \end{gathered}$$

Step 3. Find the work required to pump all of the water out of the tank.

The work required to pump water of the slice is:

$dw=91045\mathrm{\pi y}∆\mathrm{y}$

$6 feet=1.83 meter$

$$\begin{gathered} W={\int }_0^{1.83}91045\mathrm{\pi dy} \\ =91045\mathrm{\pi }{\left[\frac{{\mathrm{y}}^2}{2}\right]}_0^{1.83} \\ =255881.3\left(\frac{1.{83}^2}{2}-0\right) \\ =478694 \end{gathered}$$

So work done to pup out the water is $478694$ meter kg.