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I would like to know if my way to solve the following integral is correct: Let $I=\frac{1}{4}\int\int_Te^{x^2/4+y^2}(4-x^2-4y^2)dxdy$, with $T=\{(x,y)\in \mathbb{R^2}:0\leq x\leq 2y,\ \ x^2+4y^2 \leq 4 \}$. I made the following substitution, $\begin{cases}x=2rcos(\theta)\\\ y=rsin(\theta)\end{cases}$, so that $dxdy=2rdrd\theta$, $T=\{(r,\theta): 0\leq cos(\theta) \leq sin(\theta), \ \ r^2\leq 1 \}=\{(r,\theta): \theta \in [\pi/4,\pi/2], \ \ r\in[0,1] \}$, thus we can say $I=\frac{1}{4}\int_{\pi/4}^{\pi/2}\int_0^1e^{r^2}(4-4r^2)2rdrd\theta=\frac{\pi}{2}\int_0^1e^{r^2}(r-r^3)dr=\frac{\pi}{2}(\frac{e-2}{2})$. Is my reasoning correct?

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  • $\begingroup$ It looks fine to me. $\endgroup$ Dec 28, 2021 at 20:07

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