Double integral of $\exp(x^2+y^2)$ over the upper semi circle of radius $r=1$ 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?
 A: 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
$$
\displaystyle\mathop{\int\!\!\int}_{\substack{x^2+y^2\leq 1\\ y\geq 0}}
e^{x^2+y^2} 
\mathrm{d}\, A
=
\displaystyle\mathop{\int\!\!\int}
_{\substack{0\leq r\leq 1\\ 0 \leq\theta \leq \pi}}
r\cdot e^{r^2}
\mathrm{d}\, A
=
\int_{0}^{1}\int_{0}^{+\pi}
r\cdot e^{r^2}
\mathrm{d}\,\theta\,
\mathrm{d}\,r
$$
