Considering an appropriate mapping $(x, y)=T(u,v)$ I want to calculate $\iint_D(x^2+y^2)dxdy$ by changing the variables, where $D$ is the space that is on the first quadrant of$xy$-plane and between the curves with equations $xy=1$, $xy=3$, $x^2-y^2=1$ and $x^2-y^2=4$.
The mapping should map $D$ to a rectangle $E$ in the $uv$-plane with sides parallel to the axes.
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I have done the following:
We define new variables with the following formulas: $$u=xy, \ v=x^2-y^2$$
Then we have to solve for $x,y$ as a function of $u,v$.
$$u=xy\Rightarrow x=\frac{u}{y}$$ $$v=x^2-y^2\Rightarrow v=\left (\frac{u}{y}\right )^2-y^2\Rightarrow v=\frac{u^2}{y^2}-y^2\Rightarrow vy^2=u^2-y^4 \Rightarrow (y^2)^2+vy^2-u^2=0$$ Setting $y^2=:m$ we get $$m^2+vm-u^2=0\Rightarrow m_{1,2}=\frac{-v\pm \sqrt{v^2+4u^2}}{2}$$ Therefore we get $$y^2=\frac{-v\pm \sqrt{v^2+4u^2}}{2}$$ Since $\frac{-v-\sqrt{v^2+4u^2}}{2}<0$ it follows that $$y^2=\frac{-v+ \sqrt{v^2+4u^2}}{2}\Rightarrow y=\pm \sqrt{\frac{-v+ \sqrt{v^2+4u^2}}{2}}$$ Since $D$ is on the first quadrant, it must hold that $x\geq 0, y\geq 0$.
So, we get that $$y= \sqrt{\frac{-v+ \sqrt{v^2+4u^2}}{2}}$$ and therefore $$x=\frac{u}{y}=\frac{u}{ \sqrt{\frac{-v+ \sqrt{v^2+4u^2}}{2}}}=\frac{\sqrt{2}u}{ \sqrt{-v+ \sqrt{v^2+4u^2}}}$$
Is everything correct so far?
Do we have to calculate now the following? $$\frac{\partial{(x,y)}}{\partial{(u,v)}}$$
$$\frac{\partial{(u,v)}}{\partial{(x,y)}}=-2(x^2+y^2)$$ then $${\bf J}=\frac{\partial{(x,y)}}{\partial{(u,v)}}=\dfrac{1}{\frac{\partial{(u,v)}}{\partial{(x,y)}}}=\dfrac{1}{-2(x^2+y^2)}$$ so $(x^2+y^2){\bf J}=\dfrac{1}{-2}$ and $$\iint_D(x^2+y^2)dxdy=\iint_{D'}(x^2+y^2){\bf J}dudv=\int_1^3\int_1^4\dfrac{1}{-2}dudv$$