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I'm looking for a way calculate the following integral: $$\iint_D\frac{(x-y)^2(1+2y)}{(1+x+y^2} d(x,y)$$ With $D=\{(x,y)\in \mathbb{R}^2 : 0 \leq x+y^2 \leq 4 \mbox{ and } x\leq y\leq x+2\}$.

what I have: change of variables: $u = x+y^2$ and $v = y-x$. So $D^*$ becomes: $D^*=\{(u,v)\in \mathbb{R}^2 : 0 \leq u\leq 4, 0\leq v \leq 2 \}$. The Jacobian I calculated is $\frac{1}{(y+1)^2}$. But now comes the nasty part: rewriting the integral in terms of u and v. What I have so far: $$ \iint_{D^*}\frac{(v)^2(1+2y)}{(1+u)} \frac{1}{(y+1)^2} d(u,v) $$

Continuing any simplifications I get stuck. I tried combinations like $uv$ or $uu$, but still I'm getting no where. Could anyone help me or get in me in the right directions? Maybe a more handy choice of change of variables? Or a hint for better manipulation?

Many thanks in advance!

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No, you computed the Jacobian wrongly. $$J = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} =\begin{vmatrix} 1 & 2y\\ -1 & 1 \end{vmatrix} = 1 + 2y$$ Therefore the original integration becomes $$\iint_{D^*}\frac{v^2(1+2y)}{1+u} \frac{1}{1+2y} \,du\,dv = \iint_{D^*}\frac{v^2}{1+u} \,du\,dv$$

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  • $\begingroup$ Thanks for pointing out my mistake in calculating the Jacobian. In my book I found that for the transformation we get: d(x,y) = |diff(x,y)/diff(u,v)| d(u,v), so in this case would the jacobian not be 1/(1+2y) and thus, the term in the integral cancel out? $\endgroup$ – tymo57 May 3 '15 at 15:10
  • $\begingroup$ Replace $dxdy$ by $Jdudv$, for example when we convert cartesian to polar, $J=r$. You can just google it. $\endgroup$ – MonkeyKing May 3 '15 at 15:16
  • $\begingroup$ Sorry, I'm doubting you, really appreciate your help. But, that is my point, actually. In the polar case we take $x = r \cos(\phi)$ and $y = r \sin(\phi)$. in which case we get $\frac{\delta (x,y)}{\delta (r,\phi)} = r$ but now we don't have the transformation in the form $x=x(u,v)$ and $y=y(u,v)$, but the other way around (that's why there is coming out something in terms of $(x,y)$ and not $(r,\phi)$ or $(u,v)$). So with u= u(x,y) and v=v(x,y) would have $\frac{\delta(u,v)}{\delta(x,y)}$, but we need the inverse, therefore the inverse? Do you get my point? (Or see the flaw in my reasoning?) $\endgroup$ – tymo57 May 3 '15 at 15:33
  • $\begingroup$ Oh silly me, I was doing $u=u(x,y), v=v(x,y)$! Not realizing it for long time. I'm so bad... By the way you help me and thanks! :) $\endgroup$ – MonkeyKing May 3 '15 at 16:35
  • $\begingroup$ No problem, I cannot upvote yet, but my gratitude goes beyond that! ;-) $\endgroup$ – tymo57 May 3 '15 at 17:44

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