Answer
$e-1$
Work Step by Step
By Fubini's Theorem,
$$\int_{0}^{1}\int_{0}^{1}e^{\max\{x^2,y^2\}}dydx=\iint_R e^{\max\{x^2,y^2\}}dA$$
Here,
$$R=\{(x,y):0\leq x\leq 1,0\leq y\leq 1\}$$
We can divide $R$ into the triangles $T_1=\{(x,y):0\leq x\leq 1,0\leq y\leq x\}$ and $T_2=\{(x,y):0\leq y\leq 1,0\leq x\leq y\}$. It follows that
$$\iint_R e^{\max\{x^2,y^2\}}dA=\iint_{T_1} e^{\max\{x^2,y^2\}}dA+\iint_{T_2} e^{\max\{x^2,y^2\}}dA$$
Notice that
$$(x,y)\in T_1 \Rightarrow x\geq y\geq 0 \Rightarrow x^2\geq y^2\Rightarrow \max\{x^2,y^2\}=x^2$$
Similarly,
$$(x,y)\in T_2 \Rightarrow y\geq x\geq 0 \Rightarrow y^2\geq x^2\Rightarrow \max\{x^2,y^2\}=y^2$$
It follows that
$$\iint_{T_1} e^{\max\{x^2,y^2\}}dA+\iint_{T_2} e^{\max\{x^2,y^2\}}dA= \iint_{T_1} e^{x^2}dA+\iint_{T_2} e^{y^2}dA$$
$$=\int_{0}^{1}\int_{0}^{x}e^{x^2}dydx+\int_{0}^{1}\int_{0}^{y}e^{y^2}dxdy=\int_{0}^{1}e^{x^2}\int_{0}^{x}dydx+\int_{0}^{1}e^{y^2}\int_{0}^{y}dxdy$$
$$=\int_{0}^{1}xe^{x^2}dx+\int_{0}^{1}ye^{y^2}dy=\int_{0}^{1}2xe^{x^2}dx=e^{1^2}-e^{0^2}=e-1$$
Thus,
$$\int_{0}^{1}\int_{0}^{1}e^{\max\{x^2,y^2\}}dydx=e-1$$
