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Where D is the domain given by: $ D:=\{(x,y) : x \geq 0, -1 \leq y \leq 2, 4 \leq x^2 + y ^2 \leq 16 , 1\leq x^2 - y^2 \leq 9\} $

I'm not sure what change of variables to use here.

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    The inside hyperbola starts out properly inside the inner circle, so for negative $y$ and some positive $y$ the inner hyperbola is not the left bound of the region, I think..2012-11-04

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It doesn't look like there's a good change of variables, and your integral setup needs some care: The inner circle $x^2+y^2=4$ crosses the inner ellipse $x^2-y^2=1$ at the point $(\sqrt{5/2},\sqrt{3/2})$. The outer circle $x^2+y^2=16$ crosses the outer ellipse $x^2-y^2=9$ at the point $(\sqrt{25/2},\sqrt{7/2})$. (The other crossings of ellipses with hyperbolas are irrelevant in setting up your region.)

So it seems you'll have to break up the $y$ range into

[1] $-1, where you use the outer ellipse and inner circle as bounds,

[2] $\sqrt{3/2}, where you use the outer and inner ellipses as bounds,

[3] $\sqrt{7/2}, where you use the outer circle and inner ellipse as bounds.

That is, using line segments parallel to the $x$ axis to define your region, the formulas for the endpoints of the segments change, accrding to whether $y$ is in one of the cases [1], [2], [3] above. So it seems unlikely there's a slicker way to set things up. Of course that's only a "hunch" -- maybe there is a clever variable change.

EDIT: When I put the integral into maple using the bounds above, the value (if I entered things right) came out $(11/4)\ln{2}-\ln{3}+1/4=1.057542...$ Maybe someone could verify this value by doing the integral numerically. I also noticed that the integral over what I called case 3 was negative; I haven't checked as to whether the integrand is negative in that region.