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

Answer:

$Average = 4411.4ft^3$

Step-by-step 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:

$Time = \frac{Volume}{Rate}$

$Time = \frac{141.39ft^3}{1ft^3/min}$

$Time = 141.39\ min$

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:

$Average = \frac{1}{2} * 8822.736\ lb$

$Average = 4411.368\ lb$

$Average = 4411.4ft^3$  

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:

$V = \frac{1}{2}t^2 |\limits^{141.39}_0$

Divide by t

$\frac{V}{t} = \frac{1}{2}t |\limits^{141.39}_0$

$\frac{V}{t} = \frac{1}{2}(141.39-0)$

$\frac{V}{t} = \frac{1}{2}(141.39)$

$\frac{V}{t} = \frac{1}{2}*141.39$

$\frac{V}{t} = 70.695$

Average Volume =

$Average = \frac{V}{t} * Density$

$Average = 70.695 * 62.4$

$Average = 4411.368$

$Average = 4411.4ft^3$

The calculated value of average volume in both cases is:

$Average = 4411.4ft^3$