## Thomas' Calculus 13th Edition

$36$
Re-write as: $\int_C (x^2+y^2) \ dy=\int_{C_{1}} (x^2+y^2) \ dy+\int_{C_{2}} (x^2+y^2) \ dy ....(1)$ For $C_1$: $\int_{C_{1}} (x^2+y^2) \ dy=\int_{C_{1}} (x^2+(0)^2) \ (0)=0$ For $C_2$: $\int_{C_{2}} (x^2+y^2) \ dy=\int_{0}^{3} (3^2+y^2) \ dy=[9y +\dfrac{y^3}{3}]_0^3=(9)(3)+\dfrac{3^3}{3}=36$ Equation (1) becomes: $\int_C (x^2+y^2) \ dy=\int_{C_{1}} (x^2+y^2) \ dy+\int_{C_{2}} (x^2+y^2) \ dy =0+36=36$