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Math Help - double integral, conditions

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by fiksi View Post
    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.
    Last edited by Jester; March 16th 2009 at 01:52 PM. Reason: typo
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  3. #3
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    Quote Originally Posted by danny arrigo View Post
    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
    r2 instead of this.

    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?
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  4. #4
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    Quote Originally Posted by fiksi View Post
    Yes, we have in sqrt 1 + x2 + y2, not just x2 + y2. But i see now, i can put
    r2 instead of this.

    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?
    I fixed the typo and yes, that's the answer I got.
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  5. #5
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    Quote Originally Posted by danny arrigo View Post
    I fixed the typo and yes, that's the answer I got.
    thx... i think that is it then. Seems u know ur way around this, can u maybe provide any insight into my other integral here?

    Anyway, thx once again.
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