Advanced Calculus: How can I find the surface area of the portion of a sphere that is within a paraboloid

How can I find the surface area of the portion of a sphere (given as x^2+y^2+z^2 = 3*c^2) that is within the paraboloid 2c*z = x^2+y^2? Here, c is a positive constant. This problem can be solved both with cartesian (z,y,z) and with spherical (rho, theta, phi) coordinates.

Using Cartesian Coordinates:

Note that z = (x^2 + y^2)/(2c) intersects the sphere when
x^2 + y^2 + [(x^2 + y^2)/(2c)]^2 = 3c^2
==> (4c^2) (x^2 + y^2) + (x^2 + y^2)^2 = 12c^4
==> (x^2 + y^2)^2 + (4c^2) (x^2 + y^2) - 12c^4 = 0
==> [(x^2 + y^2) + 6c^2] [(x^2 + y^2) - 2c^2] = 0
==> x^2 + y^2 = 2c^2 (for real solutions).
This yields the region of integration.

So, the surface area equals
∫∫ √[1 + (z_x)^2 + (z_y)^2] dA
= ∫∫ √[1 + (-x/√(3c^2 - x^2 - y^2))^2 + (-y/√(3c^2 - x^2 - y^2))^2] dA
= ∫∫ √[1 + (x^2 + y^2)/(3c^2 - x^2 - y^2)] dA
= ∫∫ √[3 / (3c^2 - x^2 - y^2)] dA

Now, use polar coordinates.
Since x^2 + y^2 = 2c^2 ==> r = c√2, we obtain
∫(θ = 0 to 2π) ∫(r = 0 to c√2) √[3 / (3c^2 - r^2)] * (r dr dθ)
= 2π√3 ∫(r = 0 to c√2) r dr/√(3c^2 - r^2)
= π√3 * -2√(3c^2 - r^2) {for r = 0 to c√2}
= 2π√3 * [-√(3c^2 - 2c^2) + √(3c^2)]
= 2πc(3 - √3).

I hope this helps!