# How to integrate this?

1. Jul 19, 2012

### 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.

2. Jul 19, 2012

### 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.

3. Jul 19, 2012

### tiny-tim

hi aruwin!
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

Last edited: Jul 19, 2012
4. Jul 19, 2012

### aruwin

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.