# Double Integral of two concentric circles

1. Apr 20, 2008

### robbondo

1. The problem statement, all variables and given/known data
Let D be the region given as the set of (x,y) where 1 <! x^2+y^2 <! 2 and y !<0. Is D an elementary region? Evaluate $$\int\int_{D} f(x,y) dA$$ where f(x,y) = 1+xy.

2. Relevant equations

3. The attempt at a solution

So I understand that this is two concentric circles(an elementary region) which I can break down into two halves. So what I attempted was to break it into two times the first integral
of x from 2^(1/2) to 0 and the y from (1-x^2)^(1/2) to (2-x^2)^(1/2) so for my final solution I got (pi + 2)/4 the correct solution was pi/2 so I looked up the solution and they used the limit of x from 1 to 2^(1/2). That didn't make a whole lot of sense to me because x actually goes from 0 to 2^(1/2) doesn't it? Can someone attempt to explain why they chose their limits of integration the way they did? Is my method correct and maybe I just screwed up during the stupid Trig. Substitutions which is also very likely. Thanks~

2. Apr 20, 2008

### EngageEngage

I would do this in polar coordinates, where dA = rdrdtheta, this should make it easier to do the limits of integration and the integration itself. I'm not sure what y!<0 means..., but you r limit on r will be from 1 to sqrt(2)

3. Apr 20, 2008

### rootX

I think it can be because they might have different limits from y.

y from (1-x^2)^(1/2) to (2-x^2)^(1/2), and from 0 to (2-x^2)^(1/2)

Why don't you use cylindrical co-od!?

4. Apr 20, 2008

### robbondo

Sorry I meant 1 <= x^2+y^2 <= 2 and y <= 0. So even if I switch to different coordinate system the main conceptual issue I'm having is why you take the integral of x from square root of two to 1 instead of square root of two to 0.

5. Apr 20, 2008

### EngageEngage

if you went from square root of two to 0 you would get the entire part of the circle whose radius spans from the origin to sqrt(2). to change this, you integrate from 1 to sqrt(2). this way you will have a circle with a hole in it that has radius 1. that is what you want, righ?

6. Apr 20, 2008

### robbondo

I thought that because the limits of integration for y were from (1-x^2)^1/2 to (2-x^2)^1/2 would prevent that from happening...

7. Apr 20, 2008

### EngageEngage

Ahh, i see what you're saying, I didn't notice that and was thinking in polar.

Edit, but still x will be from 1 to sqrt(2). You will see this if you draw a picture.

8. Apr 21, 2008

### robbondo

I guess I'm still a little dissatisfied with the reasoning as to why the limits of integration of x go from 1 to root 2. I asked my TA this afternoon and she assured me that I should use the integral from -root two to root two for x. Now though If I don't use symmetry I'm going to have to break the integral up into two parts. I was told that I cannot use symmetry for this problem because the function f(x) may not be symmetric. I feel like this is something I should have in my book, but it's nowhere to be found. All the example programs deal with solid shapes, not anything like this problem. The weird thing is that the solution I have does use symmetry and does have x from 1 to root 2 and they get the correct answer, so i'm sure that's right but WHY... So far all I've heard is to use polar coordinates which doesn't really anwer my basic question...

9. Apr 21, 2008

### rootX

what are their y-limits?

this may help
http://www.ekmpowershop1.com/ekmps/shops/tyrrell123/images/doughnut.gif
I see the way you are thinking.
I think x will go from 1 to sqrt 2
only if you take y as radius of many co-centric circles
and find volume for each tiny area ..
and those co-centric circles go from 1 to sqrt 2

In that dough nut, try to think of many circle with very small intervals, and radius y