calculus 3 math i am having troubles?

Convert to polar coordinates.
z = 7 - √(x^2+y^2) ==> z = 7 - r.
x^2 + y^2 = 6x ==> r^2 = 6r cos θ ==> r = 6 cos θ, which is completely traced out with θ in [0, π].

So, the volume equals
∫∫ [7 - √(x^2+y^2)] dA
= ∫(θ = 0 to π) ∫(r = 0 to 6 cos θ) (7 - r) * (r dr dθ), via polar coordinates
= ∫(θ = 0 to π) ∫(r = 0 to 6 cos θ) (7r - r^2) dr dθ
= ∫(θ = 0 to π) [(7/2)r^2 - (1/3)r^3] {for r = 0 to 6 cos θ} dθ
= ∫(θ = 0 to π) [126 cos^2(θ) - 72 cos^3(θ)] dθ
= ∫(θ = 0 to π) [126 * (1/2)(1 + cos(2θ)) - 72 (1 - sin^2(θ)) cos θ] dθ
= ∫(θ = 0 to π) [63(1 + cos(2θ)) - 72(1 - sin^2(θ)) cos θ] dθ
= [63(θ + (1/2)sin(2θ)) - 72(sin θ - (1/3)sin^3(θ))] {for θ = 0 to π}
= 63π.

I hope this helps!