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I am attempting a question that asks me to evaluate $$ \iint_D (x^2+y^2) dx dy $$ where $D$ is the finite region in the positive quadrant bounded by the curves $$ x^2 - y^2 = \pm 1, \quad xy = \frac{1}{2}. $$

So, I thought of using the transformation $x^2 - y^2 = u$ and $xy=v$, and computed the Jacobian and all that and get to the integral

$$ \iint_D \frac{1}{v} du dv. $$

But I can't get the bounds to work. I thought $u$ would be between $-1$ and $1$, and $v$ would be between $0$ and $\frac{1}{2}$. However then the integral is not finite. Thoughts?

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I think your Jacobian is not right. I get $$ \frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} 2x & y \\ -2y & x \end{vmatrix} = 2(x^2 + y^2) $$ In terms of the area elements, $$ du\,dv = 2(x^2+y^2)\,dx\,dy $$ So if $R$ is the rectangle $[-1,1] \times [0,\frac{1}{2}]$ in the $uv$-plane, then $$ \iint_D (x^2+y^2)\,dx\,dy = \frac{1}{2}\iint_R \,du \,dv = \frac{1}{2} \cdot \operatorname{Area}(R) = \frac{1}{2}. $$

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    Yes I had gone through defining $x$ and $y$ in terms of $u,v$, and must have gotten lost in the transformation. Thank you ! – 2017-02-12