With the help of a suitable transformation and Fubini I want to determine the integral $$ \int_{V} x^{3} y d \lambda_{2}(x, y), $$ where $V$ is the open subset of $\mathbb{R}_{+}^{2}$ bounded by the following curves: $$ \begin{aligned} &x^{2}+y^{2}=4\\ &x^{2}-y^{2}=2 \\ &x^{2}-y^{2}=1 \end{aligned} $$ I know how to do that. The only problem is finding $V.$ Is it $$ V=\{(x,y)\in\mathbb{R}^2: 1<x^{2}-y^{2}<2, 0< x^{2}+y^{2}<4\} $$ Because then I set $$ \begin{aligned} &x^{2}+y^{2}=v\\ &x^{2}-y^{2}=u \end{aligned} $$ and get $$ (x,y)=\left(\sqrt{1/2(v-u}),\sqrt{1/2(u+v)}\right) $$ So either my transformation is wrong or the limits of the intervalls I chose.
Thanks for any kind of help.
We have $$ V=\left\{(x, y) \in \mathbb{R}_+^{2}: x,y>0, x^2+y^2<4, 1<x^2-y^2<2\right\} . $$ Let us consider the new variables $$ u=x^2+y^2, \quad v=x^2-y^2 $$ and the corresponding figure $$ (u, v)=\Psi(x, y):=\left(x^2+y^2, x^2-y^2\right) . $$ Obviously we have $$ U:=\Psi(V)=\left\{(u, v) \in \mathbb{R}_+^{2}: 1<v<2,v<u<4\right\} . $$ The mapping $\Psi: V \rightarrow U$ is continuously differentiable and the inverse mapping $\Phi=\Psi^{-1}$ exists and looks like this: $$ (x, y)=\Phi(u, v)=\left(\sqrt{\frac{u+v}{2}},\sqrt{\frac{u-v}{2}}\right) . $$ The mapping $\Phi: U \rightarrow V$ is also continuously differentiable and hence a diffeomorphism. We have $$ \Phi^{\prime}=\left(\begin{array}{cc} \partial_{u} \Phi_{1} & \partial_{v} \Phi_{1} \\ \partial_{u} \Phi_{2} & \partial_{v} \Phi_{2} \end{array}\right) =\frac{1}{2\sqrt{2}}\left(\begin{array}{cc} \dfrac{1}{\sqrt{u+v}} & \dfrac{1}{\sqrt{u+v}} \\ \dfrac{1}{\sqrt{u-v}} & -\dfrac{1}{\sqrt{u-v}} \end{array}\right) $$ and $$ \det (\Phi^{\prime})=-\frac{1}{u} \cdot \frac{1}{\sqrt{u^2-v^2}} . $$ Using the transformation theorem and Fubini's theorem, we obtain $$ \begin{aligned} &\int_{V} x^{3} y \hspace{1mm} d \lambda_{2}(x, y)=\int_{U} x^{3}y \hspace{1mm} |\det (\Phi^{\prime})|\hspace{1mm} d \lambda_{2}(u, v)= \int_{U} \frac{1}{4} \left(\sqrt{\frac{u+v}{2}}\right)^{3}\sqrt{\frac{u-v}{2}}\frac{1}{\sqrt{u^2-v^2}} \hspace{1mm} d \lambda_{2}(u, v)\\ &=\int_1^2\left(\int_v^4 \dfrac{\sqrt{u-v}\left(u+v\right)^\frac{3}{2}}{16\sqrt{u^2-v^2}} du\right) dv =\int_{1}^{2}\left[- \dfrac{3v^2-8v-16}{32}\right]_{v}^{4} d v = \left[-\dfrac{v\left(v^2-4v-16\right)}{32}\right]_1^{2}=\dfrac{21}{32} \end{aligned} $$