Find the value of the integral $\int_0^1\int_0^1e^{\max(x^2,y^2)}dxdy$

You have to break up the integral into 2 pieces according to which side of the line $y=x$ you are on:

$$\int_0^1 dx \, \int_0^1 dy \, e^{\max{(x^2,y^2)}} = \int_0^1 dx \, \int_0^x dy \, e^{x^2} + \int_0^1 dx \, \int_x^1 dy \, e^{y^2}$$

By reversing the order of integration in the second integral, you can show that the two pieces are equal; thus the integral is simply

$$2 \int_0^1 dx \, x \, e^{x^2} = e-1$$

Series is not the way you want to solve this problem, although it is possible. When you examine $\max(x^2,y^2)$ over the region $[0,1]\times[0,1]$, you find that: $$\max(x^2,y^2) = \begin{cases} y^2, \text{ for } y \ge x \\ x^2, \text{ for } y \lt x \end{cases}$$ Thus, by symmetry:

$$\begin{align} \int_0^1\int_0^1e^{\max(x^2,y^2)}dy\,dx &= 2\int_0^1\int_0^xe^{x^2}dy\,dx \\ &= 2\int_0^1xe^{x^2}dx\\ &= \ldots \end{align}$$ (Which is certainly integrable with Calc II methods)

Your attempt to split into cases is the right instinct, but you didn't execute it correctly. When $x\ge y$, we have $\max(x^2,y^2)=x^2$. By symmetry, we get $$\int_0^1\int_0^1 e^{\max(x^2,y^2)}\,dy\,dx = 2 \int_0^1\int_0^x e^{x^2}\,dy\,dx = \int_0^1 2xe^{x^2}\,dx = e-1\,.$$