I need help solving a triple integral

Question

∫2π to 0 ∫ (√3)/2 to 0 ∫ √(1-r²) to 1/2 [2z-1] dz r dr dθ

Answer 1

inside integral:

[ z^2 - z ], z= 1-r^2 to z = 1/2

(1/4 - 1/2) - ( 1-r^2)^2 - (1-r^2) =

(-1/4) - (1-r^2) [ (1-r^2) - 1] =

(-1/4) - (1-r^2) [ -r^2] =

(-1/4) + r^2(1-r^2) =

(-1/4) + r^2 - r^4 =

r^4 + r^2 - 1/4

2nd integral

(1/5)r^5 + (1/3)r^3 - 1/4r, r = sqrt(3)/2 to r = 0

(1/4) (sqrt(3)/2)^5 + (1/3) (sqrt(3)/2)^3 - (1/4)(Sqrt(3)/2) =

(1/4) (9sqrt(3)/32) + (1/3) ( 3 sqrt(3)/8) - (1/4)(Sqrt(3)/2) =

9sqrt(3)/128 + sqrt(3)/8 - sqrt(3)/8 =

9sqrt(3)/128

it is being subtracting from zero, so this integral evaluates to

-9sqrt(3)/128

integrating:

- 9 sqrt(3)/128 * T

9*Sqrt(3)/128 * (2*pi)