Compute the integral of f(x,y,z) = (y-x)z over the region within the cylinder x^2+y^2=16 between the planes z=0 and z=y.
Work and explanation much appreciated, thanks! :DD
Applying cylindrical coordinates yields
∫∫∫ (y - x)z dV
= ∫(θ = 0 to 2π) ∫(r = 0 to 4) ∫(z = 0 to r sin θ) (r sin θ - r cos θ) z * (r dz dr dθ)
= ∫(θ = 0 to 2π) ∫(r = 0 to 4) ∫(z = 0 to r sin θ) zr^2 (sin θ - cos θ) dz dr dθ
= ∫(θ = 0 to 2π) ∫(r = 0 to 4) (1/2) z^2 r^2 (sin θ - cos θ) {for z = 0 to r sin θ} dr dθ
= ∫(θ = 0 to 2π) ∫(r = 0 to 4) (1/2) (r^2 sin^2(θ)) * r^2 (sin θ - cos θ) dr dθ
= ∫(θ = 0 to 2π) (sin^3(θ) - sin^2(θ) cos θ) dθ * ∫(r = 0 to 4) (1/2) r^4 dr
= ∫(θ = 0 to 2π) [(1 - cos^2(θ)) sin θ - sin^2(θ) cos θ] dθ * ∫(r = 0 to 4) (1/2) r^4 dr
= 0, due to the first factor.
I hope this helps!
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Verified answer
Applying cylindrical coordinates yields
∫∫∫ (y - x)z dV
= ∫(θ = 0 to 2π) ∫(r = 0 to 4) ∫(z = 0 to r sin θ) (r sin θ - r cos θ) z * (r dz dr dθ)
= ∫(θ = 0 to 2π) ∫(r = 0 to 4) ∫(z = 0 to r sin θ) zr^2 (sin θ - cos θ) dz dr dθ
= ∫(θ = 0 to 2π) ∫(r = 0 to 4) (1/2) z^2 r^2 (sin θ - cos θ) {for z = 0 to r sin θ} dr dθ
= ∫(θ = 0 to 2π) ∫(r = 0 to 4) (1/2) (r^2 sin^2(θ)) * r^2 (sin θ - cos θ) dr dθ
= ∫(θ = 0 to 2π) (sin^3(θ) - sin^2(θ) cos θ) dθ * ∫(r = 0 to 4) (1/2) r^4 dr
= ∫(θ = 0 to 2π) [(1 - cos^2(θ)) sin θ - sin^2(θ) cos θ] dθ * ∫(r = 0 to 4) (1/2) r^4 dr
= 0, due to the first factor.
I hope this helps!