Prove $\iint_{x^{2}+y^{2} \leq 1} e^{x^{2}+y^{2}} d x d y \leq\left[\int_{-\frac{\sqrt{\pi}}{2}}^{\frac{\sqrt{\pi}}{2}} e^{x^{2}} d x\right]^{2} $ The formula we need to prove is
$$
\displaystyle \iint_{x^{2}+y^{2} \leq 1} e^{x^{2}+y^{2}} d x d y \leq\left[\int_{-\frac{\sqrt{\pi}}{2}}^{\frac{\sqrt{\pi}}{2}} e^{x^{2}} d x\right]^{2} 
$$
As I already done the left side
$$
\iint_{x^{2}+y^{2} \leq 1} e^{x^{2}+y^{2}} d x d y  = \int_{-\pi}^{\pi} \mathrm{d}\varphi \int_{0}^{1} e^{r^2}r\mathrm{d}r  = 2\pi \int_{0}^{1} e^{r^2}r\mathrm{d}r = \left.\pi e^{r^2}\right|_{0}^{1} = \pi(e-1)
$$
I still have no idea about the right side.
 A: Try to prove $$\int_{x^{2}+y^{2}\leq 1}e^{x^{2}+y^{2}}\mathrm{d}x\mathrm{d}y\leq\int_{\vert x\vert,\vert y\vert\leq\frac{\sqrt{\pi}}{2}}e^{x^{2}+y^{2}}\mathrm{d}x\mathrm{d}y$$
By considering the area of two integral domain and function value on them.
A: Integral on the left is the integration over the $\color{red}{red}$ circle, and the integral on the right is the integration over the $\color{blue}{blue}$ square (see the picture below).
We note that the area of the circle is $S_\circ = \pi\cdot 1^2 = \pi$ and the area of the square is $S_\square = (\sqrt \pi)^2 = \pi$. This means the $\color{green}{green}$ and $\color{orange}{orange}$ areas are equal too.
However, note that $$e^{\color{green}{x^2+y^2}} \le e^{\color{orange}{x^2+y^2}} $$ since $\color{green}{x^2+y^2} \le 1$ and $\color{orange}{x^2+y^2} \ge 1$ and therefore
$$\begin{align}
\iint\limits_{x^2+y^2\le 1} e^{x^2+y^2} dxdy 
&\le \iint\limits_{\text{square}} e^{x^2+y^2}dxdy \\
&= \int_{-\sqrt\pi/2}^{\sqrt\pi/2}e^{x^2}dx\int_{-\sqrt\pi/2}^{\sqrt\pi/2}e^{y^2}dy \\
&= \left(\int_{-\sqrt\pi/2}^{\sqrt\pi/2}e^{x^2}dx\right)^2
\end{align}$$

$\qquad\qquad\qquad$
A: 
In case that you don't understand what @Sqr mean

$$
\iint_{x^2+y^2}
e^{x^2+y^2\leq 1}\mathrm{d}
x\mathrm{d} y
\qquad
\iint_{|x|,|y|\leq \frac{\sqrt{\pi}}{2}} e^{x^2+y^2}\mathrm{d} x\mathrm{d} y
$$
The domain $D_1=\{x^2+y^2\leq 1\}$ and $D_2=\{|x|,|y|\leq \frac{\sqrt {\pi} }{2}\}$ have same area $\pi 1^2={\left(2\frac{\sqrt \pi}{2}\right)}^2=\pi$
Consider the picture, where
***111***
00----00
0/    \0
1|    |1
0\    /0
00----00
***111***

where 1s are in $D_1$ and 0s are in $D_2$,
we should have #1=#0.
Now let's inspect the difference between the two integral.
The 1s have $\sqrt{x^2+y^2}\leq 1$ while the 0s have $\sqrt{x^2+y^2}\geq 1$.
So the integral of $e^{x^2+y^2}$ on $D_2-D_1$ is greater than the one on $D_1-D_2$.
Therefore $\iint_{D_1} f\mathrm{d}\sigma\leq \iint_{D_2}f\mathrm{d}\sigma$
