Answer:
0
Step-by-step explanation:
∫∫8xydA
converting to polar coordinates, x = rcosθ and y = rsinθ and dA = rdrdθ.
So,
∫∫8xydA = ∫∫8(rcosθ)(rsinθ)rdrdθ = ∫∫8r²(cosθsinθ)rdrdθ = ∫∫8r³(cosθsinθ)drdθ
So we integrate r from 0 to 9 and θ from 0 to 2π.
∫∫8r³(cosθsinθ)drdθ = 8∫[∫r³dr](cosθsinθ)dθ
= 8∫[r⁴/4]₀⁹(cosθsinθ)dθ
= 8∫[9⁴/4 - 0⁴/4](cosθsinθ)dθ
= 8[6561/4]∫(cosθsinθ)dθ
= 13122∫(cosθsinθ)dθ
Since sin2θ = 2sinθcosθ, sinθcosθ = (sin2θ)/2
Substituting this we have
13122∫(cosθsinθ)dθ = 13122∫(1/2)(sin2θ)dθ
= 13122/2[-cos2θ]/2 from 0 to 2π
13122/2[-cos2θ]/2 = 13122/4[-cos2(2π) - cos2(0)]
= -13122/4[cos4π - cos(0)]
= -13122/4[1 - 1]
= -13122/4 × 0
= 0
