Double integral of √(x² + y²) over unit square using change of variables

[ninjacookies]

I'm trying to evaluate the double integral

$$\int \int \sqrt{x^2 + y^2} \, dA$$

over the region R = [0,1] x [0,1]
using change of variables.

Well, after fooling around, I've got an answer. I set u = x^2, v =y^2, and then calculated the jacobian of T which was 1. The image transformation limits of integration for u and v turned out to be the same [0,1] x [0,1]

So I did the following calculation (both integrals going from 0 to 1)

$$\int \int \sqrt{u + v} * (1) dudv$$


which resulted in a value of roughly 3.238.

Does my logic and answer seem sound here? Thanks in advance.

 

[PowerWill]

Converting to rectangular coordinates would probably be easier

 

[PowerWill]

That's cylindrical coordinates, sorry

 

[hypermorphism]

Answered in Calculus and Analysis.

 

[ninjacookies]

I'm totally at a loss here guys. I realized my Jacobian was computed wrong. Can someone please give me a clue as to what would be the most efficient integral setup? I'm completely dumbfounded. :( Thanks


edit: more in-depth post in the calculus forum, thanks

 

[dextercioby]

Please,DO NOT DOUBLE POST!

Daniel.

 

[PowerWill]

Dexter...you need to drop the intensity down a notch. And to the ninja, just convert $$x^2 + y^2$$ to $$r^2$$ and integrate over the same area in cylindrical.

 

[dextercioby]

He can't do that,the square [0,1]*[0,1] is not equivalent to the quarter of the circle you're implying...

Daniel.