Answer
$36$
Work Step by Step
We can write the integral as follows:
$\int_C (x^2+y^2) \ dy=\int_{C_{1}} (x^2+y^2) \ dy+\int_{C_{2}} (x^2+y^2) \ dy ....(1)$
$\int_{C_{1}} (x^2+y^2) \ dy=\int_{C_{1}} (x^2+(0)^2) \ (0)=0 ...(2)$
$\int_{C_{2}} (x^2+y^2) \ dy=\int_{0}^{3} (3^2+y^2) \ dy=[9y +\dfrac{y^3}{3}]_0^3=(9)(3)+\dfrac{3^3}{3}=36 ...(3)$
Using equations (2) and (3), equation (1) becomes:
$\int_C (x^2+y^2) \ dy=\int_{C_{1}} (x^2+y^2) \ dy+\int_{C_{2}} (x^2+y^2) \ dy \\ =0+36 \\=36$
