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The question is: $$\iint_D xy \ dxdy, \quad D=1\leq x^2+y^2\leq2, \ \ \ x^2\leq y\leq x^2+1, \ x\geq0 , \ y\geq0$$ I've tried this: $$ u=x^2+y^2 , \ v= y-x^2 , \ \ |J|=\frac{1}{2x(1+2y)}$$ $$\frac{1}{2}\iint \frac{y}{1+2y} \ dudv$$ But how should i solve for $y$ in terms of $u$ and $v$? Any suggestion would be great, thanks

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  • $\begingroup$ You have to find for y in terms of u and v from equations $x^2+y^2=u$ and $y-x^2=v$. Sum them you get a quadratic equation for y interms of u and v . $\endgroup$ – sirous Jan 10 at 17:13
  • $\begingroup$ Do you mean $D=\{(x,\,y)\in(0,\,\infty)^2|1\le x^2+y^2\le2,\,x^2\le y\le x^2+1\}$ (D=\{(x,\,y)\in(0,\,\infty)^2|1\le x^2+y^2\le2,\,x^2\le y\le x^2+1\})? $\endgroup$ – J.G. Jan 10 at 17:46
  • $\begingroup$ @J.G I have updated the region, does it matter? $\endgroup$ – simon Jan 10 at 17:54
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    $\begingroup$ @simon Thanks for adding clarity. $\endgroup$ – J.G. Jan 10 at 17:56
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Since $y=v+x^2=u+v-y^2$ has unique positive root $y=\tfrac{\sqrt{4(u+v)+1}-1}{2}$, $\tfrac{y}{1+2y}=\tfrac12\left(1-\tfrac{1}{\sqrt{4(u+v)+1}}\right)$. Now we evaluate$$\int_1^2du\int_0^1dv\tfrac12\left(1-\tfrac{1}{\sqrt{4(u+v)+1}}\right)=\tfrac{1-A}{2},\,A:=\int_1^2du\int_0^1dv\tfrac{1}{\sqrt{4(u+v)+1}}.$$Modulo possible mistakes in my arithmetic you should check for,$$\begin{align}A&=\int_1^2du[\tfrac12\sqrt{4u+4v+1}]_0^1\\&=\tfrac12\int_1^2du[\sqrt{4u+5}-\sqrt{4u+1}]\\&=\tfrac{1}{12}[(4u+5)^{3/2}-(4u+1)^{3/2}]_1^2\\&=\tfrac{13^{3/2}-54+5^{3/2}}{12},\end{align}$$so your original integral was $\tfrac{11}{4}-\tfrac{13^{3/2}+5^{3/2}}{24}$.

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Unless you find a better change of variable, my suggestion is to integrate it based on the sketch. It is split into two integral and is really not that bad.

$\displaystyle \int_a^1 y \ \bigg[ \ \int_{\sqrt{1-y^2}}^{\sqrt y} x \ dx \ \bigg] \ dy $ + $\displaystyle \int_1^b y \ \bigg [ \int_{\sqrt{y-1}}^{\sqrt {2-y^2}} x \ dx \ \bigg ] \ dy$

$a$ is the value of $y$ at intersection of $y = x^2, x^2 + y^2 = 1 \implies y^2+y-1=0$. This gives you $a = \frac{\sqrt 5 - 1}{2}$

$b$ is the value of $y$ at intersection of $y = x^2 + 1, x^2 + y^2 = 2 \implies y^2+y-3=0$. This gives you $b = \frac{\sqrt 13 - 1}{2}$.

And the correct answer is ${\frac{36+5\sqrt5 - 13 \sqrt{13}}{48}}$

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    $\begingroup$ @ Math Lover , I agree with you, this is a better method of solving this problem than with change of variable $\endgroup$ – simon Jan 11 at 5:37

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