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We do have:

$\int\int_{D} \frac{xy}{y^{2}-x^{2}}dxdy$

Being D the region limited by the curves:

$x^{2}-y^{2} = 1$

$x^{2}-y^{2} = 4$

$\frac{x^{2}}{16}+\frac{y^{2}}{4} = 1$

$\frac{x^{2}}{4}+y^{2} = 1$

My attempt from the hint:

Should I solve for this integral?

$\int\int_{D} \frac{5x^{2}y^{2}}{y^{2}-x^{2}}dxdy$

How could I get the extremes of the integral from the four equations given?

Thanks

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  • $\begingroup$ Why did I receive negative feedback? I could update my question showing my attempt if need be $\endgroup$ – JD_PM May 28 '18 at 14:39
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    $\begingroup$ Yes it should be better if you show your work and effort here. $\endgroup$ – gimusi May 28 '18 at 14:39
  • $\begingroup$ @gimusi Please if you could provide more details in why you set up the integral like you did, I may understand it better. Thanks $\endgroup$ – JD_PM Jun 23 '18 at 8:59
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Make a sketch of the domain

enter image description here

and for symmetry we can deduce that the integral over the four part of the region delimited by the curves is equal to zero.

To evaluate the integral for the first quadrant let

  • $u=x^2-y^2$
  • $v=\frac{x^{2}}{4}+y^{2}$

then $$dudv=|J|dxdy=\begin{vmatrix}2x&-2y\\\frac x 2&2y\end{vmatrix}dxdy=5xy>0$$

thus

$$\int\int_{D} \frac{xy}{y^{2}-x^{2}}dxdy=2\int_1^4dv\int_1^4 -\frac1udu=-6\ln 4$$

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  • $\begingroup$ Could you provide more details in your solution please? I am not getting your answer $\endgroup$ – JD_PM May 28 '18 at 17:43

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