Calculus 2- Hooke's Law Spring problem

[1LastTry]

Calculus 2- Water pump problem

Homework Statement

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.

Homework Equations

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?

 

[vela]

You forgot the attachment.

 

[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?

 

[1LastTry]

You forgot the attachment.

Sorry forgot, there it is.

 

[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)

 

[1LastTry]

My BAD!

my bad I got the title wrong.

Well here is another attachment...

So the volume of the disk is \pi(4-x^2)dx (\pir^{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?