# integrating in polar co-ordinates

• Apr 14th 2007, 11:53 PM
elbarto
integrating in polar co-ordinates
Hi,
Just wondering if someone could please have a look at my working for the question i am currently working on. I got a bit of help from CaptainBlack on this forum about a week or so about how to approch the problem and this is what i have come up with so far. I was reading through my notes and noticed they talk of and use a factor called the jacobian. does this factor have to be used everytime you integrate in polar co-ordinates? If so how would i apply it to this problem. I get a negative for my answer which is not what i was expecting.
Thanks
Elbarto
• Apr 15th 2007, 12:26 AM
Jhevon
Quote:

Originally Posted by elbarto
Hi,
Just wondering if someone could please have a look at my working for the question i am currently working on. I got a bit of help from CaptainBlack on this forum about a week or so about how to approch the problem and this is what i have come up with so far. I was reading through my notes and noticed they talk of and use a factor called the jacobian. does this factor have to be used everytime you integrate in polar co-ordinates? If so how would i apply it to this problem. I get a negative for my answer which is not what i was expecting.
Thanks
Elbarto

Seems ok to me
• Apr 15th 2007, 04:28 AM
topsquark
Quote:

Originally Posted by elbarto
I get a negative for my answer which is not what i was expecting.
Thanks
Elbarto

First, you would expect it to be negative. After all, if x^2 y^2 < 1 and then the argument of ln is always less than 1, the range of which is, in fact, negative.

However I see a problem. The expression u*ln(u) - u is not defined at u = 0. (So neither is your integral.) As it happens, the limit of u*ln(u) as u --> 0 is 0, but that is not what your integral is evaluating. Your domain of integration would need to be:
0< x^2 + y^2 <= 1

-Dan
• Apr 15th 2007, 04:56 AM
elbarto
I think i understand what you mean about x^2 y^2 < 1 giving a negative answer. I assume this would be the case up until x^2 + y^2 >= e^2?

Are you saying that my answer is correct but only through coincidence that the limit of u*ln(u) is 0?
How does 0< x^2 + y^2 <= 1 effect the limits i used for my integration. is 0<r<1 and 0<theata<2pi still the area of integration i am dealing with? Sorry for all the question i am just finding the concept of integrating over areas (particularly areas expressed as polars) a little different to the general integration i have done in the past.

Thanks Alot
Elbarto