Verified answer

Let I = ∫(-∞ to ∞) e^(-x^2) dx.

So, I^2 = ∫(-∞ to ∞) e^(-x^2) dx * ∫(-∞ to ∞) e^(-y^2) dy

...........= ∫(-∞ to ∞) ∫(-∞ to ∞) e^(-x^2-y^2) dy dx.

Converting to polar coordinates yields

I^2 = ∫(θ = 0 to 2π) ∫(r = 0 to ∞) e^(-r^2) * (r dr dθ); note that there are no cosines in sight.

.....= ∫(θ = 0 to 2π) dθ ∫(r = 0 to ∞) r e^(-r^2) dr

.....= 2π * (-1/2)e^(-r^2) {for r = 0 to ∞}

.....= π.

Since I > 0, taking square roots yields I = √π.

I hope this helps!

it is not likely a double fundamental - it is one fundamental subtracted from yet another - outdoors certain minus the interior certain: first element is you're staring on the 1st quadrant, which ought to offer you the bounds of integration from 0 to ?/2. A = a million/2 ?r1^2 d? - a million/2 ?r2^2d?