# Calculus 2- Hooke's Law Spring problem

1. May 25, 2013

### 1LastTry

Calculus 2- Water pump problem

1. The problem statement, all variables and given/known data

The lower half of a sphere with radius 2m is filled with water. Find the work needed to pump the water to a point 1m above the centre of the sphere.

2. Relevant equations

3. The attempt at a solution
I know you have to cut the problem into small pieces, and you have to label a "x" anywhere you want but you have to solve the problem based on which part you label the x (see attachment). But I have no idea where to label the x and how to get the integral with that information. And also what is the limits I put in the definite integral? I know it has to do with the x you choose but how to know which limit to use?

#### Attached Files:

• ###### Calc2-2.png
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Last edited: May 25, 2013
2. May 25, 2013

### vela

Staff Emeritus
You forgot the attachment.

3. May 25, 2013

### Simon Bridge

If you didn't have to worry about which part you'd label "x", then how would you go about it?
Wouldn't you define a coordinate system first - exploiting the symmetry of the situation?

Note: it would take an infinite work to pump all the water into a point - since water does not readily compress that far.

Perhaps you just need to lift the water to a height 1m above the initial level of the water?

4. May 25, 2013

### 1LastTry

Sorry forgot, there it is.

5. May 25, 2013

### Simon Bridge

Cool - looks like x is decided for you.
Isn't it the position of the disk with thickness dx in the diagram?

I also don't see what this has to do with Hook's Law.

Anyway - it sounds to me that you are trying to apply an equation you do not fully understand.

Start by defining a coordinate system. You can put the origin anywhere you like - so use your knowledge of geometry and the symmetry of the system to put it where it makes the math easier.

Notice that a disk of thickness dx at height x has mass dm (which depends on the density of water and the volume of the disk)...

The amount of work you need to lift a mass dm to height h from height x is dW. dW=... (you finish)

Last edited: May 25, 2013
6. May 25, 2013

### 1LastTry

my bad I got the title wrong.

Well here is another attachment...

So the volume of the disk is $\pi$(4-x^2)dx ($\pi$r$^{2}$h)
dw = $\pi$(4-x^2)dx x 1000 x 9.8 (density of water and gravitational acceleration) x dx

Wait...work is force times distance, what would be the distance be? The distance from dx to the 1m mark above the bowl? so x+1? So the Work (W) would be dW * (1+x)?

And the integral would be that as well? and the limits of the definite integral would be 0 to 2?

#### Attached Files:

• ###### Calc2-2.png
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Views:
81
Last edited: May 25, 2013