Question

Water is run at a constant rate of ft^3/min to fill a cylindrical tank of radius ft and height ft. Assuming that the tank is initially empty, make a conjecture about the average weight of the water in the tank over the time period required to fill it, and then check your conjecture by integrating. [Take the weight density of water to be lb/ft^3.]

Carry out all calculations exactly then round the final answer to one decimal place.

Answer 1

`Average = 4411.4ft^3`

Explanation:

Given [Missing from the question]

`Rate = 1ft^3/min`

`Radius = 3ft`

`Height = 5ft`

`Density = 62.4lb/ft^3`

The volume (V) of a cylindrical tank is:

`V = \pi r^2h`

This gives:

`V = \pi * 3^2 * 5`

`V = \pi * 45`

`V = 3.142 * 45`

`V = 141.39ft^3`

If the rate is given as:

`Rate = 1ft^3/min`

Then the time to fill the tank is:

`Time = (Volume)/(Rate)`

`Time = (141.39ft^3)/(1ft^3/min)`

`Time = 141.39\ min`

The weight of the water when the tank is filled up is:

`Weight = 141.39ft^3 * 62.4lb/ft^3`

`Weight = 141.39 * 62.4lb`

`Weight = 8822.736\ lb`

The conjecture about the average weight is:

`Average = (1)/(2) * 8822.736\ lb`

`Average = 4411.368\ lb`

`Average = 4411.4ft^3`

To check by integrating:

After time t:

The volume (V) of the water in tank is

`V = \int\limits^(141.39)_0 {t} \, dt`

i.e. from time = 0 to 141.39 minutes

Integrate:

`V = (1)/(2)t^2 |\limits^(141.39)_0`

Divide by t

`(V)/(t) = (1)/(2)t |\limits^(141.39)_0`

`(V)/(t) = (1)/(2)(141.39-0)`

`(V)/(t) = (1)/(2)(141.39)`

`(V)/(t) = (1)/(2)*141.39`

`(V)/(t) = 70.695`

Average Volume =

`Average = (V)/(t) * Density`

`Average = 70.695 * 62.4`

`Average = 4411.368`

`Average = 4411.4ft^3`

The calculated value of average volume in both cases is:

`Average = 4411.4ft^3`