How to Transform Limits and Factor in the Jacobian for Polar Coordinates?

[aruwin]

Given x=rcosθ, y=2rcosθ

Integrate the following:
I = ∫∫_D{x²+y²/4} dxdy

D = {(x,y)|x²+y²/4≤1}

My first attempt was that I substitute the polar coordinate given into the domain for x and y and so I got rcosθ+rsinθ≤1. And from here, I don't know what else to do,I am stuck.Help.

 

[chiro]

Hey aruwin.

If you are going to make a transformation you need to change the limits as well as factor in the Jacobian going from (x,y) to (r,theta).

Show us your working with how you did the transformation from (x,y) to (r,theta) with x=rcos(theta), y=rsin(theta) and tell us where you get stuck if you get stuck.

 

[tiny-tim]

hi aruwin!

… rcosθ+rsinθ≤1 …

nooo …

D is r²cos²θ+r²sin²θ ≤ 1, isn't it?

get some sleep! :zzz:​

EDIT: ah, seems it was a mis-type …

but you can now write that as simply r² ≤ 1

 

[aruwin]

Hey aruwin.

If you are going to make a transformation you need to change the limits as well as factor in the Jacobian going from (x,y) to (r,theta).

Show us your working with how you did the transformation from (x,y) to (r,theta) with x=rcos(theta), y=rsin(theta) and tell us where you get stuck if you get stuck.

x is already given as rcosθ and y as 2rsinθ, so they are already in their polar coordinates,right? Now I have to change the limits too and what I did was subsititutiing those x and y values in their polar forms into x²+y²/4 ≤1
So from here, I get r²cos²θ+r²sin²θ≤1.
And I square root all of them and they became
√(r²cos²θ+r²sin²θ)≤1.

From here, clearly there's a right angle triangle with hypothenuse 1. But how do I get the limits for θ? And for r, I guess it's pretty obvious that it's limit is from 0 to 1. And that's all I was able to do,if it's correct.