First, understand that your given integral is a constant (sqrt(2)) times the integral over a certain integral domain of dy.dx
That is to say, S = sqrt(2)*Area of the domain.
What is the domain ?
x is in range [-6 ; 6], and x being chosen, y is in range [ - sqrt(36-x²) ; sqrt(36-x²) ].
This domain is therefore the disk of center (0;0) and radius 6.
Its area is : pi*radius ² = pi * 6² = 36*pi
So finally, you got S = sqrt(2)*36*pi
dx has not been changed to dThêta, and dy has not been changed to rdr.
Actually, dx.dy has been changed to r.dr.dThêta, according to a new parametrizing of the integration domain :
Thêta is in range [0 ; 2*pi], while r is in range [ 0;6].
