Question

Help with jacobian http://postimg.org/image/yutmuzm7h/

Answer 1

i reckon that you should do a little algebra first

so you have \(y = \mu \omega\) and \(x + y = \mu\)

from that you can get \(x = \mu - \mu \omega\)

i think you are good to go from there

Answer 2

I'm doing good? http://postimg.org/image/a8mwkhe1h/ Thanks for your answer :D

Answer 3

i think the J is 1

i haven't looked at it otherwise but i can try if that would be helpful

Answer 4

I will be happy if you do help me with this ;_; and I think that J is 1 indeed

Answer 5

Help please I got stucked and I don't know If I did good with my steps of the image from top D:

Answer 6

omg what is that Irishboy123? D:

Answer 7

Those are the boundaries of the limit.

Answer 8

That is the answer then? I don't get it I just want to know the answer if it's easy why nobody tells me :(

Answer 9

naru

no-one said its easy :p

first of all , go back check the jacobian. i was in a rush last night, now have a bit more time to work this if you are around.... i think \(J = \mu\) but do it for yourself to be sure...

that drawing is the area you are integrating over on the (x,y) system

now we have a transformation, we will have a new area.

as we have \(x = \mu - \mu \omega\) and \(y = \mu \omega\) we can start to draw a region in \((\mu, \omega )\) co-ords



for the line \(y = 1-x\) we can say that \(\mu \omega = 1 - (\mu - \mu \omega)\) meaning \(\mu \omega = 1 - \mu + \mu \omega\) so \(\omega = 1\)

can you do this for the other 2 lines that mark out the region ie , y = 0 amd x = 0??

i'll have a look at the work you posted too...

hope this is of help.

Answer 10

soz!! big typo...

i repeat: "for the line y=1−x we can say that \(\mu \omega=1−(\mu−\mu \omega)\) meaning \(\mu \omega=1−\mu+\mu \omega\) so \(\color{red}{\mu=1}\)"

so we have the first line of our new integration region..