Convert the integral $$I = \int_0^{4/\sqrt{2}} \int_y^{\sqrt{16 - y^2}} e^{2x^{2}+2y^{2}} \,dx\,dy$$ to polar coordinates, getting $$\int_C^D \int_A^B h(r,\theta)\,dr\,d\theta,$$ where

$h(r,\theta) =$

$A =$

$B =$

$C =$

$D =$

and then evaluate the resulting integral to get

$I =$ .