I'm really not sure on formatting so I'll give it my best shot, but my working out at the moment leaves me at the point of
$$\int_0^1\int_0^{\pi} e^{r^2} rd\theta dr$$
So between $\pi$ and $0$ for the left integral (because it's a semi circle not a circle) and between $1$ and $0$ for the right integral.
Would this be correct?
We have, for $x(r,\theta)=r\cos\theta$ e $y(r,\theta)=r\sin(\theta)$ $$ \displaystyle\mathop{\int\!\!\int}_{\substack{x^2+y^2\leq 1\\ y\geq 0}} e^{x^2+y^2} \mathrm{d}\, A = \displaystyle\mathop{\int\!\!\int} _{\substack{r^2(x,y)\leq 1\\ r(x,y)\cdot \sin (\theta(x,y))\geq 0}} e^{x^2(r,\theta)+y^2(r,\theta)} \mathrm{d}\, A = \displaystyle\mathop{\int\!\!\int} _{\substack{0\leq r\leq 1\\ -\pi \leq\theta \leq \pi}} e^{x^2(r,\theta)+y^2(r,\theta)} \left|\det\frac{\partial(x,y)}{\partial(r,\varphi)}\right| \mathrm{d}\, A. $$ Observation: we have $y=r\sin \theta\geq 0$ if, only if, $\sin \theta \geq 0$ by cause $r>0$. And $\cos \theta \geq 0$ if, only if, $0\leq \theta \leq +\pi$.
Recall that $$ J = \left|\det\frac{\partial(x,y)}{\partial(r,\theta)}\right| =\begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\[8pt] \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} =\begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} =r\cos^2\theta + r\sin^2\theta = r. $$ Then $$ \displaystyle\mathop{\int\!\!\int}_{\substack{x^2+y^2\leq 1\\ y\geq 0}} e^{x^2+y^2} \mathrm{d}\, A = \displaystyle\mathop{\int\!\!\int} _{\substack{0\leq r\leq 1\\ 0 \leq\theta \leq \pi}} r\cdot e^{r^2} \mathrm{d}\, A = \int_{0}^{1}\int_{0}^{+\pi} r\cdot e^{r^2} \mathrm{d}\,\theta\, \mathrm{d}\,r $$