Find ∫∫∫ sinzdv, where Ω is surrounded by a cone z = root (x ^ 2 + y ^ 2) and a plane y = π

Find ∫∫∫ sinzdv, where Ω is surrounded by a cone z = root (x ^ 2 + y ^ 2) and a plane y = π

The cross section equation is D: x ^ 2 + y ^ 2 = Z ^ 2, the cross section is a circle, its area is: π Z ^ 2 ∫ ∫ sinzdv = ∫ Sinz (∫ ∫ DXDY) DZ, the integral area of the double integral in the middle is cross section D, because the integrand is 1, the result is cross section area =