# Thread: average value word problem

1. ## average value word problem

Water is run at a constant rate of 1 ft^3/min to fill a cylindrical tank of radius 3ft and height 5ft. 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 62.4 lb/ft^3).

i found the volume of the cylinder to be 235.619 ft^3 which means it will take 235.619 min to fill the tank. so i set up the integral as (1/235.619) integral from (0 to 235.619) of (62.4) dt. i didn't get the correct answer. the correct answer is 1404(pi). please help me set up the correct integral.

2. The volume of the tank is:
$
V = \pi R^2h = \pi(3 ft)^2*5ft = 45 \pi ft^3
$

Since the tank fills at a constant rate of $1 ft^3/min$, it takes the tank $45\pi$ min to fill, and the weight of water as a function of time is:

$
W = \int _0 ^T ( 1 ft^3/min * 62.4 lb/ft^3) dt = T*62.4 lb/min.
$

The "average" weight is simply the weight when the tank is half way through the time required to fill it, which is at $T = \frac 1 2 45 \pi \ min$:
$
W = \frac 1 2 * 45 \pi \ min * 62.4\ lb/min = 1404 \pi \ lbs
$

3. how come after you take the integral you don't divide by 45pi? isn't the formula for average value the integral divided by (b-a) which in this case is (T-0) = T = 45pi?

4. If we did this in a more formal way it would be like this:

$
W_{Avg} = \frac 1 T \int _{t=a} ^{t=b} w(t)*dt
$

For your problem the weight as a function of time is $w(t) = \rho R t$, where $\rho$ is the density of water = 62.4 lb/ft^3, and R is the fill rate = 1 ft^3/min. The limits on the integral are a = 0 min and b = 45 $\pi$ min. So:

$
W_Avg = \frac 1 {45 \pi} \int _ 0 ^{45 \pi} 62.4 \frac {lb} {ft^3} * 1 \frac {ft^3} {min} t dt = \frac 1 {2* 45 \pi} 62.4 * (45 \pi) ^ 2 = \frac 1 2 62.4 *45 \pi = 1404 \pi \ lb
$