How do you solve this double integral

How do you solve the double integral if f(x,y) = x
and the region you are integrating it over is the region in the first quadrant that lies between x^2+y^2 = 64 and x^2+y^2 = 8x

i know you want to change this to polar coordinates, the double integral is gonna be r^2cos(Θ), and iknow that the bounds for dΘ is gonna be 0 to π/2, but i can't figure out what the bounds are going to be for dr? Can somebody walk me through how to solve this double integral?

Note that converting the equations of the circles to polar coordinates yields
x^2 + y^2 = 64 ==> r = 8.
x^2 + y^2 = 8x ==> r^2 = 8r cos θ ==> r = 8 cos θ.

So, ∫∫ x dA
= ∫(θ = 0 to π/2) ∫(r = 8 cos θ to 8) (r cos θ) * (r dr dθ)
= ∫(θ = 0 to π/2) (1/3) r^3 cos θ {for r = 8 cos θ to 8} dθ
= ∫(θ = 0 to π/2) (512/3) (1 - cos^3(θ)) cos θ dθ
= (512/3) ∫(θ = 0 to π/2) [cos θ - cos^4(θ)] dθ
= (512/3) ∫(θ = 0 to π/2) [cos θ - ((1/2)(1 + cos(2θ))^2] dθ
= (512/3) ∫(θ = 0 to π/2) [cos θ - (1/4)(1 + 2 cos(2θ) + cos^2(2θ))] dθ
= (128/3) ∫(θ = 0 to π/2) [4 cos θ - (1 + 2 cos(2θ) + (1/2)(1 + cos(4θ)))] dθ
= (64/3) ∫(θ = 0 to π/2) [8 cos θ - (2 + 4 cos(2θ) + (1 + cos(4θ)))] dθ
= (64/3) ∫(θ = 0 to π/2) [8 cos θ - 3 - 4 cos(2θ) - cos(4θ)] dθ
= (64/3) [8 sin θ - 3θ - 2 sin(2θ) - sin(4θ)/4] {for θ = 0 to π/2}
= (64/3) (8 - 3π/2)
= (32/3) (16 - 3π).

I hope this helps!