\int_0^1\!\int_0^1 e^{\max\{x^2,y^2\}}\,dy\,dx.
Solution. Let $R=\{(x,y)\mid 0\le x,y \le 1\}$. For $x,y\in R$, $\max\{x^2,y^2\} = x^2$ if $x\ge y$ and $\max\{x^2,y^2\}=y^2$ if $x\le y$. Therefore, we divide $R$ into two regions: $R=R_1 \cup R_2$,
where $R_1=\{(x,y)\mid 0\le
x\le 1, 0\le y\le x\}$ and $R_2=\{(x,y)\mid 0\le y\le 1, 0\le x\le
y\}$.
Thus $\max\{x^2,y^2\}=x^2$ for $(x,y)\in R_1$ and $\max\{x^2,y^2\}=y^2$ for $(x,y)\in R_2$. This implies
\begin{eqnarray}
\int_0^1\!\int_0^1 e^{\max\{x^2,y^2\}}\,dy\,dx &= &
\int\!\!\int_R
e^{\max\{x^2,y^2\}}\,dA \\
&=& \int\!\!\int_{R_1} e^{\max\{x^2,y^2\}}\,dA +
\int\!\!\int_{R_2}
e^{\max\{x^2,y^2\}}\,dA \\
&=&\int_0^1\!\int_0^x e^{x^2}\,dy\,dx + \int_0^1\!\int_0^y
e^{y^2}\,dx\,dy \\
&=&\int_0^1 xe^{x^2}\,dx + \int_0^1 ye^{y^2}\,dy \\
&=& e^{x^2} \Big |_0^1 = e-1.
\end{eqnarray}