Defining $u=x^2+y^2$, $v=x^2+y^2-2y$, I want to calculate $\iint_D xe^y dxdy$ using the change of variable, where $D$is the space that is determined by the inequalities $x\geq 0$, $y\geq 0$, $x^2+y^2\leq 1$ and $x^2+y^2\geq 2y$.
$$$$
I have done the following:
We have the new variables $u=x^2+y^2$, $v=x^2+y^2-2y$.
When we solve for $x,y$ as a function of $u,v$ we get: $$v=x^2+y^2-2y\Rightarrow v=u-2y \Rightarrow y=\frac{u}{2}-\frac{v}{2}$$ $$u=x^2+y^2 \Rightarrow x^2=u-y^2 \Rightarrow x^2=u-\left (\frac{u-v}{2}\right )^2 \Rightarrow x^2=u-\frac{u^2-2uv+v^2}{4} \\ \Rightarrow x^2=\frac{4u-u^2+2uv-v^2}{4} \Rightarrow x=\pm \sqrt{\frac{4u-u^2+2uv-v^2}{4}}\\ \Rightarrow x=\pm \frac{\sqrt{4u-u^2+2uv-v^2}}{2}$$ Since $x\geq 0$ we get $x= \frac{\sqrt{4u-u^2+2uv-v^2}}{2}$.
Is everything correct so far?
Now we have to calculate the determinant, right? We get \begin{align*}det \begin{pmatrix}\frac{\partial{x}}{\partial{u}} & \frac{\partial{x}}{\partial{v}} \\ \frac{\partial{y}}{\partial{u}} & \frac{\partial{y}}{\partial{v}}\end{pmatrix}&=det \begin{pmatrix}\frac{2-u+v}{2\sqrt{4u-u^2+2uv-v^2}} & \frac{u-v}{2\sqrt{4u-u^2+2uv-v^2}} \\ \frac{1}{2} & -\frac{1}{2}\end{pmatrix} \\ & =-\frac{2-u+v}{4\sqrt{4u-u^2+2uv-v^2}}-\frac{u-v}{4\sqrt{4u-u^2+2uv-v^2}} \\ & =-\frac{2-u+v+u-v}{4\sqrt{4u-u^2+2uv-v^2}} \\ & =-\frac{1}{2\sqrt{4u-u^2+2uv-v^2}}\end{align*}
Therefore we get the following: \begin{align*}\iint_D xe^y dxdy&=\iint_E \frac{\sqrt{4u-u^2+2uv-v^2}}{2}e^{\frac{u-v}{2}} \left |\frac{\partial{(x,y)}}{\partial{(u,v)}}\right |dudv \\ & =\iint_E \frac{\sqrt{4u-u^2+2uv-v^2}}{2}e^{\frac{u-v}{2}}\frac{1}{2\sqrt{4u-u^2+2uv-v^2}}dudv \\ & =\frac{1}{4}\iint_E e^{\frac{u-v}{2}}dudv\end{align*}
Is the new space $E=\{(u,v) \mid 0\leq u\leq 1, 0\leq v\leq 1\}$ ?
I thought so because of the following:
Since $x^2+y^2\leq 1 \Rightarrow u\leq 1$. Since $x\geq 0, y\geq 0$ we have that $x^2+y^2\geq 0$ and so $u\geq 0$.
Since $x^2+y^2\geq 2y \Rightarrow v\geq 0$. Since $x^2+y^2\leq 1$ and $y\geq 0$ we have that $x^2+y^2-2y\leq 1-2\cdot 0=1$ and so $0\leq v\leq 1$.