Double integral of $xe^y$ over the area inside $x^2 + y^2 = 1$ but outside $x^2 + y^2 = 2y$

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My question goes like this:

Let R be the area inside $x^2 + y^2 = 1$ and outside $x^2 + y^2 = 2y$. Calculate $\int\int_R xe^y dA$.

How sould I approach this question? I tried to use integration with polar cordinates, but then I end up with the following very complicated integral when inserting $x = r \cos\theta$ and $y = r \sin \theta$:

$$\int_0^{\frac{\pi}{4}} \int_1^{2\sin\theta} r^2 \cos\theta e^{r \sin\theta} dr d\theta$$

Is this a wrong approach, or have I simply done something wrong? Should i use substitution of variables instead? If yes, any suggestions as to which substitution?

All help is very much apprectiated!

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Notice that $R$ is a region that is symmetrical about the $y$-axis. For every point $(x, y)\in R$, we have $(-x, y) \in R$.

That is for any point that is evaluated to be $xe^y$, there is a value that is being evaluated to be $-xe^{-y}$.

The double integral is evaluated to be $0$.

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