
Yefim S. answered 01/14/22
Math Tutor with Experience
∫∫S yz dσ = ∫∫Dy(y + 3)√1 + 1dxdy = √2∫∫D(y2 + 3y)dxdy. where D = {x2 + y2 ≤ 1}.
In polar coordinates: y = rsinθ; dxdy = rdrdθ; D = {0 ≤ r ≤ 2π, 0 ≤ θ ≤ 2π}.
∫∫S yz dσ = √2∫∫D(r2sin2θ + 3rsinθ)rdrdθ = √2∫02π∫01(r3sin2θ + 3r2sinθ)drdθ = √2∫02π(r4/4sin2θ + r3sinθ)01dθ =
√2∫02π(1/8 - 1/8cos2θ + sinθ)dθ = √2(θ/8 - 1/16sin2θ - cosθ)02π = √2(π/4 - cos2π + cos0) = π√2/'4