# How to split D to calculate $\int_Df(x,y)d(x,y)$ for $f(x,y)=xe^{x^2+y^2}$

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?

• You'll probably be better off using polar coordinates – NL1992 Aug 12 at 15:23
• Also, $f$ maps $R^2$ to $R$, not to $R^2$. – JG123 Aug 12 at 15:29

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?

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