Q.54 - 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 an upright conical tank that is 12 feet high and has a radius of 6 feet at the top. 

Step-by-Step Solution

Answer

The work required to pump all of the water out of the top of an upright conical tank that is 12 feet high and has a radius of 6 feet at the top is 

$26956.8\mathrm{\pi }$

Step 1. Given information.

We have given,

The work required to pump all of the water out of the top of an upright conical tank that is 12 feet high and has a radius of 6 feet at the top. 

Step 2. Concept used.

Formula for work is, 

$W=Fd=\omega Vd$

Similar triangles:

If two triangles are similar, then their corresponding sides are in equal proportion.

Step 3. Explanation.

We have given,

The work required to pump all of the water out of the top of an upright conical tank that is 12 feet high and has a radius of 6 feet at the top.  

From the given information,

Suppose that height of the slice is y ft and radius of the height is x ft. 

Using similar triangle property,

$$\begin{gathered} \frac{x}{y}=\frac{6}{12} \\ x=\frac{1}{2}y \end{gathered}$$

The vertical distance 12 - y.

Volume of the slice =  ${\mathrm{\pi r}}^2\mathrm{dy}$

The weight density of water = $\omega =62.4$pounds per cubic ft.

Using formula for work,

$W=62.4\times \mathrm{\pi }\times {\left(\frac{1}{2}\mathrm{y}\right)}^2\times \mathrm{dy}\times (12-\mathrm{y})$

Hence, 

$$\begin{gathered} W={\int }_0^{12}62.4\times \mathrm{\pi }\times {\left(\frac{1}{2}\mathrm{y}\right)}^2\times \mathrm{dy}\times (12-\mathrm{y}) \\ =15.6\mathrm{\pi }{\int }_0^{12}{\mathrm{y}}^2(12-\mathrm{y})\mathrm{dy} \\ =15.6\mathrm{\pi }\left({\int }_0^{12}12{\mathrm{y}}^2-{\mathrm{y}}^3\mathrm{dy}\right) \\ =15.6\mathrm{\pi }\left(12{\left[\frac{{\mathrm{y}}^3}{3}\right]}_0^{12}-{\left[\frac{{\mathrm{y}}^4}{4}\right]}_0^{12}\right) \\ =15.6\mathrm{\pi }\left(6912-5184\right) \\ =26956.8\mathrm{\pi } \end{gathered}$$

Step 4. Conclusion.

Hence, 

The work required to pump all of the water out of the top of an upright conical tank that is 12 feet high and has a radius of 6 feet at the top is,

$26956.8\mathrm{\pi }$