# Applications of Integration

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• Jul 11th 2008, 10:13 AM
JonathanEyoon
Applications of Integration
I have the more difficult time with word problems. Can someone help guide me through a couple of them? More than wanting to find what the answer is, if someone could help me understand the why and how of the problems I would appreciate it.

The diameter of a circular pool is 24 m, the sides
are 5 m high, and the depth of the water is 4 m. How
much work is required to pump all of the water out
over the side?

Ok so far I've drawn a diagram which is a cylinder with radius of 12 and height of 5 meters. I've gone ahead and added the water height of 4 meters as well. Now I know I need to find how much work will be involved in getting all the water out. How would I go about this. (Speechless)
• Jul 11th 2008, 10:23 AM
galactus
The area of the kth layer of water is $\displaystyle {\pi}(12)^{2}$ with height

$\displaystyle {\Delta}y$

So, we have $\displaystyle 144{\pi}{\Delta}y$ for the volume of the kth disc.

The force required to move the kth layer equals the weight of the layer,

which can be found by multiplying its volume by the weight density of

water, which is $\displaystyle 9810 \;\ N/m^{3}$.

So, we have $\displaystyle 1412640{\pi}\int_{0}^{4}\underbrace{(5-y)}_{\text{distance moved}}dy$
• Jul 11th 2008, 10:31 AM
JonathanEyoon
whoa. From the looks of it, you did it so much simpler than how the professor did it. Could you break that down a little? I don't understand how you got everything. Sorry for being a pain.
• Jul 11th 2008, 10:40 AM
galactus
I thought I did, sorry. Since the pool is circular, you know each slice of water will have area $\displaystyle {\pi}r^{2}$. OK?.

But each 'slice' is, in effect, a very thin cylinder with height dy.

So, each has volume $\displaystyle 144{\pi}dy$.

Now, add them all up. That is where the integration comes in.

The distance each slice is pumped is 5-y, because y is the distance from the top to the kth slice.

Therefore, the distance is 5-y and the force is $\displaystyle 1412640{\pi}dy$

See?.
• Jul 11th 2008, 10:48 AM
JonathanEyoon
Ok I think I see a little of what you're saying now. Question. My professor used gravity as a factor. Why wasn't gravity used in this one?

Also, If y = the distance from the disc to the top of the tank instead of the distance from the disc to the bottom of the tank, would the boundaries be different when integrating? If so why?
• Jul 11th 2008, 11:04 AM
galactus
No, that looks good to me. You're integrating from 0 to 4 because that is how deep the water is.

Your professor probably used Newtons which is the force required to give a mass of 1 gram an acceleration of $\displaystyle 1 \;\ m/sec^{2}$

I believe you will have to multiplied by $\displaystyle 9810 \;\ N/m^{3}$

Which is the weight density of water.
• Jul 11th 2008, 11:12 AM
CaptainBlack
Quote:

Originally Posted by galactus
The area of the kth layer of water is $\displaystyle {\pi}(12)^{2}$ with height

$\displaystyle {\Delta}y$

So, we have $\displaystyle 144{\pi}{\Delta}y$ for the volume of the kth disc.

The force required to move the kth layer equals the weight of the layer,

which can be found by multiplying its volume by the weight density of

water, which is $\displaystyle 1000 \;\ kg/m^{3}$.

So, we have $\displaystyle 144000{\pi}\int_{0}^{4}\underbrace{(5-y)}_{\text{distance moved}}dy$

And the units of this work are? They should be Joules (Newton metres), but they are not because there is a factor of g missing.

Doing it this way is perpetuating the abominations incorporated in the imperial and US customary unit systems.

This is also easier to do by considering the change in potentail energy, which gives exactly this calculation multipled by g.

(The given force is not sufficent to move the water, but sufficient to support it against gravity)

RonL
• Jul 11th 2008, 11:13 AM
JonathanEyoon
ah ic. That's why gravity was used. Question. The professor told us that we should find the radius in terms of r(y) How would I do this in this problem?
• Jul 11th 2008, 11:14 AM
galactus
You are certainly correct. I forgot to multiply by 9.8 which gives us the 9810.
• Jul 11th 2008, 11:30 AM
galactus
You know, I am up in air about something now myself. Despite having done umpteen of these problems. Why is it in some calc book they just multiply by 1000 as the weight density of water and other times 9810?.

Just for kicks, I went and looked up a similar problem and the book just multiplied by 1000 and not 9810.

One book says the weight density of water is 1000 kg/m^3 and another says it is 9810 N/m^3.

Why can't we just multiply by 1000?. That is what I always done before. Just as if it were in feet we would use 62.4 lb/ft^3

You all got me second guessing myself.
• Jul 11th 2008, 12:07 PM
JonathanEyoon
I didn't notice it until now but when you're integrating, why is it 5 - y? I mean I know that it represents the distance from the disc to the top but what about the distance from the disc to the bottom? The more i'm looking at what happened, the more i'm getting confused. I can't seem to understand how we can take only the distance from the disc and up and then place boundaries from 0 to 4 which is the depth of the water given the location of the disc and the distance about is smaller than 4. Did that make sense?
• Jul 11th 2008, 12:38 PM
galactus
Perhaps it would be easier for you to visualize by writing it this way:

$\displaystyle 9810\cdot 144{\pi}\int_{1}^{5}ydy$

That is easier yet and results in the same solution.

We are going from the top down instead of from the bottom up.

I see what CB is saying about the gravity thing. Makes perfect sense. I had never noticed before, but some books just multiply by 1000 and others by 9810. The 9810 makes more sense. If this were a fluid pressure problem, then we should use 1000.
But when we're pumping something against gravity we should factor it in.
• Jul 11th 2008, 01:09 PM
JonathanEyoon
MmMmm... still no light bulb going off.

Maybe this will help you understand why I can't seem to understand. This is a problem the professor did for us and told us every problem similar to this is gonna follow the same steps.

A tank has the shape of an inverted circular cone with height 10m and base radius 4m. It is filled with water to a height of 8m. Find the work required to empty the tank by pumping all of the water to the top of the tank.

Below is an exact diagram of what he gave us. Here are the steps he gave us. Find the radius of the disc in terms of y which will be r = (2/5)(10 - Y).

Now he said dwork = (dF)(distance that the disc is raised)

I'm gonna use * as an indicator of multiplication

= dm * g * y
= p * dV * y * g
= p * pi(r)^2 * dy * g * y
= 1000 * 9.8 * pi *(4/25) integrate from 2 to 10 (y( 10 - y)^2 dy)

Now i've been using this as a reference as to what to do with these types of problems. In this problem, why did he multiply the distance from the disc and up to the distance from the disc and down? What makes this problem so different from the one i originally listed? Also why are the boundaries from 2 to 10? Shouldn't it be from 0 to 8 since that is the depth of the water?
• Jul 11th 2008, 01:24 PM
CaptainBlack
Quote:

Originally Posted by JonathanEyoon
I didn't notice it until now but when you're integrating, why is it 5 - y? I mean I know that it represents the distance from the disc to the top but what about the distance from the disc to the bottom? The more i'm looking at what happened, the more i'm getting confused. I can't seem to understand how we can take only the distance from the disc and up and then place boundaries from 0 to 4 which is the depth of the water given the location of the disc and the distance about is smaller than 4. Did that make sense?

If you think of this as the work being equal to the change in potential energy when lifting the layer the only distance of importance is how far you have to lift the layer.

RonL
• Jul 11th 2008, 01:26 PM
galactus
Your professor is integrating from the top down. The water is 2 feet below the top. Therefore, the integration is from 2 to the bottom of the cone, 10.

Your professor also used similar triangles because we are dealing with a cone, therefore, the radius of the slices are changing as move along the height of the cone. With a cylindrical pool that is constant, so it is easier.

Your professor could also have set it up like so and got the same answer. This is going from the bottom up. The other way is top down.

$\displaystyle \frac{9800\pi}{3}\int_{0}^{8}\left(2y(\frac{2}{5}y )^{2}\right)dy$

This comes from the similar triangles. $\displaystyle \frac{2}{5}=\frac{r}{y}$

$\displaystyle r=\frac{2}{5}y$

From the volume of a cone we have $\displaystyle \frac{\pi}{3}(\frac{2}{5}y)^{2}y$
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