This is a 3D representation of the object which has a circular cross-section. (Acknowledgements to GeoGebra for their graphic app.)
The plane y=6 in the x-z plane cuts through the paraboloid y=3x²+z² making a circle such that 3x²+3z²=6, that is, x²+z²=2, so the radius of the circle is √2. This sets the limits for x and z to √2, with lower limits at the origin, where x=0 and z=0. We can think of the surface as a set of discs with decreasing radius from √2 down to zero. The surface area is the sum total of the outer surface areas of the discs, which in calculus is given by integration, and in this case is a double integral involving x and z.
S=∫∫√((∂y/∂x)²+(∂y/∂z)²+1)dA where S is the total surface area and dA the infinitesimal areas we are summing. ∂y/∂x=6x and ∂y/∂z=6z, so S=∫∫√(36x²+36z²+1)dxdz, because dA=dxdz.
It is more convenient to switch to polar coordinates, replacing dxdz with rdrdθ. Note that rdθ is the arc length subtended by angle dθ, so rdrdθ is the outer surface area of a disc. r=√(x²+z²). S=∫∫r√(36r²+1)drdθ. The maximum radius of the circle is √2, as established earlier, where the paraboloid meets the plane y=6, so the limits for r are [0,√2] and the limits for θ are [0,2π].
Let u=36r²+1, then du=72rdr. The limits for u are found by plugging in the limits for r into the equation for u, that is, [0,√2]→[1,73]. S= (1/72)∫∫√ududθ=(1/72)ʃ(⅔(73^(3/2)-1)dθ for θ ∊[0,2π].
So S=(π/54)(73^(3/2)-1)=36.23 square units approx.