Math Is Fun Forum

I see why you were having problems when writing the problem like that. To imagine the shape i tend to use the polar coordinates on a problem like this, because:

, hence it indicates that the xy-plane is a disc with changing radius, however a radius can not be smaller than 0 and the shape is also bounded by the the value of 2 for the radius. We haven't bound the disc from entering any of the quadrants (i hope that is the english term) of the xy-plane, so we can let it run from 0 to 2pi.
By this information I imagine it is a cone. However i still don't get the part where:

.
It is bothering me why I can't let z be an r, since I know the density = z = r. I don't know why I can't put the boundaries as I have done. So the only part I actually know about this integral is why we multiply by r (it is because it is polar coordinates).
- The entire thing just seem to give the correct result, but why? I am clueless.

Well as I wrote, I posted all the information I was given (I translated the question), so I can't give you any more information.
If it is true that I need to consider either cylindrical or spherical coordinates in 3 dimensions, then I don't have to make this exercise, because our exam won't contain any of this, however this is one of the exercises mentioned we should be able to make. - I am kind of confused myself, so what I did was trying to calculate the result through polar coordinates.

I've given you all the information that I've been given: The 2 z-boundaries and the density.
However I would think we're working with the boundary:

Well you can solve for the solution, right? So we start out by finding the solution of the characteristic equation:

We find the solutions r=3 and r=-1. Hence we get the complete solution:

By inserting the values you've been given we obtain:

and

Solving these equations we get that

and

Hence the formula we obtain is:

If we have to add A_1, A_2...A_n, the we will get either 0 or -1. If n is an odd number, then we will get -1 as an result, and 0 if n is even.

I am just wondering how all the sources I find regarding the cycle canceling claim this:
"Any cycle canceling decreases the objective function by a strictly positive amount."
(This, as far as I understand, claim that we reduce the cost of the flow after each iteration).

If there is a proof for it, then I am asking for it, however it just seems like it is implied in one of the theorems (and those theorems I have proofs for) and I am missing out on something.

Yes the triple-w should be enough.

Hmmm it's not possible to say that all the eigenvalues are distinct (I would say). Oh I forgot a very important thing: I'm looking at left eigenvectors. So the eigenvalue problem looks as the last part of this:
[math]Ax=\lambda x \Leftrightarrow x^T(A-I)^T = x^T(A^T-I^T) = 0[\math]
It doesn't matter thinking of eigenvalues, because I've proven these are the same for the left and right eigenvectors. I think I will look at what it takes to only have distinct eigenvalues (I don't have a proof that shows that all eigenvalues are distinct).
Thanks for a response

Edit: Oh didn't see your post there! Yes I have been to that page of wikipedia, since I needed this to understand the proof. I though only have looked for some understandings of the proof! It's getting late here so I will stop the search for tonight, but thanks for the fast response, didn't expect that