# Change of coordinates should not give the same result?

Let $R$ a region defined by the interior of the circle $x^2+y^2=1$ and the exterior of the circle $x^2+y^2=2y$ and $x\geq 0$, $y\geq 0$

Using next change coordinate systems to determine the region $D$ in corresponding plane that corresponds to $R$ under given change of coordinate system

a) polar coordinates $x=r\cos t$, $y=r\sin t$ to determine the region $D$ in $rt$ plane that corresponds to $R$

i.e. $$T(r,t)=(r\cos t, r\sin t)$$

b) coordinates given by $u=x^2+y^2$, $v=x^2+y^2-2y$ to determine the region $D$ in $uv$ plane that corresponds to $R$

i.e. $$T(u,v)=(\frac{\sqrt{4u-(u-v)^2}}{2}, \frac{u-v}{2})$$

How is $D$ such that $T(D)=R$

Calculate the integral $\int\int_Rxe^y \ dx \ dy$ using this change coordinate system (polar coordinates)

The integral result must be the same using either change coordinate system, right?

I'm getting a difference from a sign in both results (between a and b)

Details

a) Jacobian for polar coordinates is $J_r=r$.

I'm getting the result of the integral as follows:

$$\int\int_Rxe^y \ dx \ dy = \int_0^1 \int_0^{arcsin(r/2)} r \ cos(t) \ e^{r\ sin(t)} \cdot r \ dt \ dr \\ =\int_0^1 \left[e^{r \ sin(t)}\right]_0^{arcsin(r/2)} r \ dr \\ =\int_0^1 r \ (e^{r^2/2}-1) \ dr \\ = e^{1/2}-\frac{3}{2}$$

Details of this development in this question

b) For this, I obtain that

$$0\leq u\leq 1 \ \ , \ \ 0\leq v \leq u$$

And when I get $u$ and $v$:

$$x=\frac{\sqrt{4u-(u-v)^2}}{2} \ \ , \ \ y=\frac{u-v}{2}$$

Jacobian is $$J_{uv}=\frac{-1}{2\sqrt{4u-(u-v)^2}}$$

Which was validated with this.

So, the integral is as follows:

$$\int\int_Rxe^y \ dx \ dy = \int_0^1 \int_0^u \left(\frac{\sqrt{4u-(u-v)^2}}{2}\right) \left(\frac{-1}{2\sqrt{4u-(u-v)^2}}\right) \ e^{\frac{u-v}{2}} \ dv \ du \\ =\frac{-1}{4} \int_0^1 \int_0^u \ e^{\frac{u-v}{2}} \ dv \ du \\ =\frac{1}{2} \int_0^1 \left(e^0-e^{u/2}\right) \ du \\ = -e^{1/2}+\frac{3}{2}$$

So, why results are different? I think I'm making a mistake, but I don't find where, I double (triple, 4 times, etc) check my operations.

Any help will be appreciated

Suppose we have $X{\times} Y\sim\mathcal U(0;1){\times}(0;1)$, and apply the transform $U=-X, V=Y$ to evaluate $\mathsf E(XY)$ (for demonstration purposes).
So $U{\times}V\sim\mathcal U(-1;0){\times}(0;1)$
Firstly $\displaystyle\mathsf E(XY) =\int_0^1\int_0^1 xy\operatorname d y\operatorname d x=1/4$
Nextly $\displaystyle\mathsf E(XY)\neq \int_{-1}^0\int_0^1 (-u)v\det\begin{bmatrix}\partial (-u)/\partial u & \partial (-u)/\partial v\\\partial v/\partial u&\partial v/\partial v\end{bmatrix}\operatorname dv\operatorname d u=-1/4$