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Consider the region $D$ of the plane in the first quadrant bounded by the hyperbolas $x^2-y^2=1$, $x^2-y^2=4$ and the circles $x^2+y^2=4$ and $x^2+y^2=9$. Use change of coordinates to find the double integral $$ \iint_Dx\;dA $$ I've graphed the region, but I don't know how to find the bounds.

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  • $\begingroup$ Welcome to MSE. Please use Latex to write the Math. $\endgroup$ – nls Nov 5 '16 at 15:55
  • $\begingroup$ Hint: $x=\sqrt{\frac{u+v}{2}}$, $y=\sqrt{\frac{v-u}{2}}$. $\endgroup$ – Kuifje Nov 5 '16 at 17:04
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If you perform the change of variables \begin{cases} x=\sqrt{\frac{v+u}{2}}\\ y=\sqrt{\frac{v-u}{2}}\\ \end{cases} Your region $D$ becomes much more friendly: it is bounded by $u=1$, $u=4$, $v=4$, $v=9$, i.e. it is a rectangle in the $(u,v)$-plane.

The Jacobian of such a transformation is $\frac{1}{2(u+v)}$.

It follows that $$ \iint_D x\; dA = \int_1^4\int_4^9\sqrt{\frac{v+u}{2}}\frac{1}{2(u+v)}\;dvdu $$

I am sure you can take it from there.

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