Calling #I =int_oo^oo e^{-x^2/2}dx# we know that
#I^2 = (int_oo^oo e^{-x^2/2}dx)(int_oo^oo e^{-y^2/2}dy)# but the integrals are independent so
#I^2 = int_oo^oo int_oo^oo e^{-(x^2+y^2)/2}dx dy#
Changing to polar coordinates
#rho^2 = x^2+y^2#
#dx dy equiv rho d rho d theta#
To cover the whole plane in polar coordinates we have
#I^2= int_{rho=0}^{rho=oo}int_{theta=0}^{theta=2pi}e^{-rho^2/2} rho d rho d theta#
#I^2 = (int_{rho=0}^{rho=oo}rho e^{-rho^2/2} d rho)(int_{theta=0}^{theta=2pi} d theta) = 1 xx 2pi#
Then #I = sqrt(2pi)#
Note:
#d/(drho)e^{-rho^2/2} = -rho e^{-rho^2/2}#