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)


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


The integrals have identical magnitude and opposing directions; and the second is negative.   The area should be unsigned.

This is because you did not take the absolute value of the Jacobian determinant to ensure orientation was preserved.

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 $\displaystyle\mathsf E(XY)\neq \int_{-1}^0\int_0^1 (-u)v\det\begin{bmatrix}\partial (-u)/\partial u & \partial (-u)/\partial v\\\partial v/\partial u&\partial v/\partial v\end{bmatrix}\operatorname dv\operatorname d u=-1/4$


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