Evaluating $\frac{1}{2\pi} \iint_{\mathbb{R}^2} e^{\frac{-x^2}{2}} e^{\frac{-y^2}{2}} \, dA$ I'm trying to evaluate the double integral
$$
\frac{1}{2\pi} \iint_{\mathbb{R}^2} e^{\frac{-x^2}{2}} e^{\frac{-y^2}{2}} \, dA.
$$
Any ideas?

Hint:
$$
\frac{1}{2\pi} \iint_{\mathbb{R}^2} e^{\frac{-x^2}{2}} e^{\frac{-y^2}{2}} \, dA = \left( \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{\frac{-x^2}{2}} \, dx \right) \left( \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{\frac{-y^2}{2}} \, dy \right) = \left( \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{\frac{-x^2}{2}} \, dx \right)^2
$$
This can easily be integrated using polar coordinates (or by recognizing that it is a normal distribution, and so the integral is $\sqrt{2\pi}$).
