Problem: Compute ∬R(3x+4y2)dA for R the region with y≥0 that lies between the circles of radii 1 and 2.
Solution: We convert the integral into polar coordinates using dA=rdrdθ. Obtain:
∬R(3x+4y2)dA=∫0π∫12(3rcos(θ)+4(rsin(θ))2)rdrdθ=∫0π(7cos(θ)+15sin2(θ))dθ=215π.