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Let $$D=\{(x,y)\in \mathbb{R}^2: 1\le x^2+y^2 \le 4 ,\quad y\ge0\}$$ Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be defined as $$f(x,y)=xe^{x^2+y^2}$$

Calculate $\int_Df(x,y)d(x,y)$

I believe the order of integration must be $dxdy$. However, I can't clearly express $x$ as a function of $y$. What is the correct way to split the set $D$ so we can integrate?

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    $\begingroup$ You'll probably be better off using polar coordinates $\endgroup$ – NL1992 Aug 12 at 15:23
  • $\begingroup$ Also, $f$ maps $R^2$ to $R$, not to $R^2$. $\endgroup$ – JG123 Aug 12 at 15:29
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Using polar coordinates you have:
$x=r\cos(t)$ and $y=r\sin(t)$. Where $r \geq 0, t \in \mathbb{R}$
Now looking at points in D we realize that,because $x^2+y^2=r^2$, we must have that $$ 1 \leq r^2 \leq 4 $$ Moreover $y\geq 0$ means that $t \in [0,\pi]$.

Now under a change of variables in the intergration we need to calculate the Jacobian which turns out to be $r$ (exercise).

Thus the integration over $D$ becomes: $$ \int_0^\pi\int_1^2r^2 \cos(t)e^{r^2} drdt $$ Which is a lot simpler.
Can you finish from here?

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  • $\begingroup$ you're missing the Jacobian term here $\endgroup$ – NL1992 Aug 12 at 15:32
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    $\begingroup$ @NL1992 thanks, edited. $\endgroup$ – Matthew Aug 12 at 15:33
  • $\begingroup$ If I understood correctly that is the determinant of the jacobian because we are chaining $f$ and the polar coordinate switch? $\endgroup$ – Ruben Kruepper Aug 12 at 15:58
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    $\begingroup$ @RubenKruepper it involves the error function so may not be so simple if you haven't seen that. symbolab.com/solver/integral-calculator/… $\endgroup$ – Matthew Aug 13 at 9:55
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    $\begingroup$ @RubenKruepper also see mathworld.wolfram.com/Erfi.html $\endgroup$ – Matthew Aug 13 at 10:05

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