Here, we only need the upper hemisphere:
z = √(36 - x^2 - y^2).
Note that the cone and sphere intersect at z^2 + z^2 = 36 ==> z = √18
==> The region of integration is x^2 + y^2 + 18 = 36 ==> x^2 + y^2 = 18.
So via Cartesian Coordinates, the surface area equals
∫∫ √[1 + (z_x)^2 + (z_y)^2] dA
= ∫∫ √[1 + (-x/√(36 - x^2 - y^2))^2 + (-y/√(36 - x^2 - y^2))^2] dA
= ∫∫ √[1 + (x^2 + y^2)/(36 - x^2 - y^2)] dA
= ∫∫ √[36/(36 - x^2 - y^2)] dA
= ∫∫ 6 dA/√(36 - x^2 - y^2).
Converting to polar coordinates yields
∫(θ = 0 to 2π) ∫(r = 0 to √18) 6r dr dθ/√(36 - r^2)
= 2π ∫(r = 0 to √18) 6r(36 - r^2)^(-1/2) dr
= 2π * -3 * 2√(36 - r^2) {for r = 0 to √18}
= 12π (6 - 3√2)
= 36π (2 - √2).
I hope this helps!
