# Average value word problem

1. Jan 13, 2010

### demonelite123

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. Jan 13, 2010

### tiny-tim

Hi demonelite123!

(have a pi: π and try using the X2 tag just above the Reply box )
eek! … that's πh2r, isn't it?

3. Jan 13, 2010

### HallsofIvy

As tiny-tim points out, the volume of a cylinder of radius r and height h is $\pi r^2 h$, not $\pi r h^2$ which you appear to be using. And, of course, for a linear problem like this the "average" weight of the water will be 1/2 the full weight.

4. Jan 13, 2010

### demonelite123

OH oops. what a silly mistake to make. thanks for the reply guys!

also, if the problem is linear i know now to just take half of the total but what if the problem is not linear?

5. Jan 13, 2010

### tiny-tim

uhh? just integrate … that will give you the answer automatically.

6. Jan 13, 2010

### demonelite123

i have another question regarding the original problem.

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?

7. Jan 14, 2010

### tiny-tim

Hi demonelite123!

(what happened to that π i gave you? )

Yes, that's right … for the average, you divide tby the time, which in this case is 45π.

I suspect you had the wrong integral … what integral did you use?