Find the value of the integral $\int_0^1\int_0^1e^{\max(x^2,y^2)}dxdy$ I am currently having difficulty trying to evaluate a certain double integral.
$$\int_0^1\int_0^1e^{\max(x^2,y^2)}dxdy$$
I've broken it down into two cases,
(i) $\max(x^2,y^2)=y^2$
then $$\int_0^1\int_0^1e^{\max(x^2,y^2)}dxdy=\int_0^1e^{y^2}dy$$
(ii) $\max(x^2,y^2)=x^2$
then $$\int_0^1\int_0^1e^{\max(x^2,y^2)}dxdy=\int_0^1e^{x^2}dx$$ since the inner integral does not depend on $y$
I believe that this is correct. It is at this point that I cannot evaluate. From my understanding this integral cannot be solved by any traditional means.
By using series, I can write 
$$\int_0^1e^{x^2}dx = \int_0^1\sum_{n=0}^{\infty}\frac{x^{2n}}{n!}dx = \sum_{n=0}^{\infty}\frac{x^{(2n+1)}}{(2n+1)(n!)}|_0^1 = \sum_{n=0}^{\infty}\frac{1}{(2n+1)(n!)}$$
Would this be the correct way of doing this? Also, if it is the correct method, is that series the appropriate answer? I started to have second thoughts here so I have not shown convergence (or not), I suspect this would be the next step.
 A: 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)
A: 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
$$\int_0^1\int_0^1 e^{\max(x^2,y^2)}\,dy\,dx = 2 \int_0^1\int_0^x e^{x^2}\,dy\,dx = \int_0^1 2xe^{x^2}\,dx = e-1\,.$$
A: 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
$$2 \int_0^1 dx \, x \, e^{x^2} = e-1$$
A: \begin{eqnarray}
\int_{x=0}^1 \int_{y=0}^1 e^{\max(x^2,y^2)} dx dy &=& 2 \int_{x=0}^1 \int_{y=0}^x e^{\max(x^2,y^2)} dx dy \\
&=& 2 \int_{x=0}^1 \int_{y=0}^x e^{x^2} dx dy \\
&=& 2 \int_{x=0}^1 x e^{x^2} dx dy \\
&=& e-1
\end{eqnarray}
