Find the area of the surface. The part of the sphere x^2 + y62 + z^2 = a^2 lies within....

need help i'm on my last chance at webassign
Find the area of the surface.
The part of the sphere
x2 + y2 + z2 = a2
that lies within the cylinder
x2 + y2 = ax
and above the xy-plane

A = ∫∫ √(1 + (z_x)^2 + (z_y)^2) dA
...= ∫∫ √(1 + (x/√(a^2 - x^2 - y^2))^2 + (y/√(a^2 - x^2 - y^2))^2) dA, since z = √(a^2 - x^2 - y^2)
...= ∫∫ √(1 + (x^2 + y^2)/(a^2 - x^2 - y^2)) dA
...= ∫∫ √(a^2/(a^2 - x^2 - y^2)) dA
...= a ∫∫ dA/√(a^2 - x^2 - y^2).

In polar coordinates x^2 + y^2 = ax ==> r = a cos θ, which is completely traced out with θ in [0, π].
So, converting to polar coordinates yields
a ∫(θ = 0 to π) ∫(r = 0 to a cos θ) (r dr dθ)/√(a^2 - r^2)
= a ∫(θ = 0 to π) -√(a^2 - r^2) {for r = 0 to a cos θ} dθ
= a ∫(θ = 0 to π) (a - a sin θ) dθ
= a (aθ + a cos θ) {for θ = 0 to π}
= a² (π - 2).

I hope this helps!