1. double integral, conditions

I have to find: I I (over S) (2/(sqrt(1+x2+y2)) dx dy

S is a set of x,y such that 1<=x2 + y2 <=4 and x>=0, y>=0.

I understand that we have a circle here, actually a circle within a circle, and as i see, conditions give me only quarter of that circle, r going from 1 to 2 and having 0 to 1/2pi, right?

I guess i should work somehow with polar form, but how to actually do the integral?

thanks

2. Originally Posted by fiksi
I have to find: I I (over S) (2/(sqrt(1+x2+y2)) dx dy

S is a set of x,y such that 1<=x2 + y2 <=4 and x>=0, y>=0.

I understand that we have a circle here, actually a circle within a circle, and as i see, conditions give me only quarter of that circle, r going from 1 to 2 and having 0 to 1/2pi, right?

I guess i should work somehow with polar form, but how to actually do the integral?

thanks
If $\iint_R \frac{2}{\sqrt{1+x^2+y^2}}\,dA = \int_0^{\pi/2} \int_1^2 \frac{2r}{\sqrt{1+r^2}}\, dr d\theta$. You should be able to do this integral.

3. Originally Posted by danny arrigo
If $\iint_R \frac{2}{\sqrt{x^2+y^2}}\,dA = \int_0^{\pi/2} \int_1^2 \frac{2r}{\sqrt{1+r^2}}\, dr d\theta$. You should be able to do this integral.
Yes, we have in sqrt 1 + x2 + y2, not just x2 + y2. But i see now, i can put

I get 2/sqrt(1+r2). If I integrate as you say, using substitution, i get pi/2(2sqrt5 - 2 sqrt2). It seems I have made a mistake somehwere? Or is this it?

4. Originally Posted by fiksi
Yes, we have in sqrt 1 + x2 + y2, not just x2 + y2. But i see now, i can put