Answer to Question #344330 in Calculus for Jane

Question

5. Suppose that a water tank is shaped like a right circular cone with the tip at the bottom, and has height 10 meters and radius 2 meters at the top. If the tank is full, how much work is required to pump all the water out over the top?

Answer 1

Let's consider the circular cross-section through the tank at depth h and find its radius using similar triangles:

$\frac{2}{r}=\frac{10}{10-h},$

$r=\frac{2(10-h)}{10}=\frac{10-h}{5}.$

The aea of the cross-section: $A=\pi r^2=\frac{\pi (10-h)^2}{25}.$

The volume of the section of water at depth h:

$V=\pi r^2dh=\frac{\pi (10-h)^2}{25}dh.$

The density of the water: $\sigma=1000 kg/m^3,$ so the mass of water in the section is

$m=\sigma V=1000\frac{\pi (10-h)^2}{25}dh=40\pi (10-h)^2dh.$

To raise this mass of water we need to overcome the force of gravity $F=mg$ , $g=9.8 m/s^2.$

$F=9.8\cdot 40\pi (10-h)^2dh=392\pi(10-h)^2dh.$

We must lift this water a distance h and require a work

$W=392\pi(10-h)^2hdh.$

The total work:

$W=\int_0^{10} 392\pi(10-h)^2hdh=392\pi \int_0^{10} (10-h)^2hdh=$

$=392\pi \int_0^{10} (100h-20h^2+h^3)dh=392\pi (100\frac{h ^2}{2}-20\frac{h^3}{3}+\frac{h^4}{4})|_0^{10}=$

$=392\pi (5000 - \frac{20,000}{3}+\frac{10,000}{4})=392\pi\frac{2500}{3}=\frac{980,000}{3}\pi\approx1,026 kJ.$

Answer: 1,026 kJ.