Integrate $\iint_D\exp\{\min(x^2, y^2)\}\mathrm{d}x\mathrm{d}y$ Compute the integral:
\begin{equation}
\iint\limits_{\substack{\displaystyle 0 \leqslant x \leqslant 1\\\displaystyle 0 \leqslant y \leqslant 1}}\exp\left\{\min(x^2, y^2)\right\}\mathrm{d}x\mathrm{d}y
\end{equation}
$D$ is the rectangle with vertices $(0,0)$, $(0,1)$, $(1,0)$ and $(1,1)$ and $\min(x^2,y^2)$ is the minimum of the numbers $x^2$ and $y^2$.
I dont have a clue about how I can integrate this function. I thought of using Taylor series but I am not sure if this would work either. Can anyone help me out?

Note that the derivative of $x\mapsto e^{x^2}$ is $x\mapsto 2xe^{x^2}$, hence by symmetry along the line $x=y$
$$\begin{align} \int_0^1\int_0^1e^{\min\{x^2,y^2\}}\,\mathrm dy\,\mathrm dx &= 2\int_0^1\int_x^1e^{x^2}\,\mathrm dy\,\mathrm dx\\
&=2\int_0^1(1-x)e^{x^2}\,\mathrm dx\\
&=2\int_0^1e^{x^2}\,\mathrm dx-\int_0^12xe^{x^2}\,\mathrm dx\\
&=2\int_0^1e^{x^2}\,\mathrm dx-e+1.
\end{align}$$
Unfortunately, the remaining integral is non-elementary.

A similar integal with max instead of min is much easier:
$$\begin{align} \int_0^1\int_0^1e^{\max\{x^2,y^2\}}\,\mathrm dy\,\mathrm dx &= 2\int_0^1\int_0^xe^{x^2}\,\mathrm dy\,\mathrm dx\\
&=2\int_0^1 x e^{x^2}\,\mathrm dx\\
&=e-1.
\end{align}$$
