## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 16: Integrals and Vector Fields - Section 16.2 - Vector Fields and Line Integrals: Work, Circulation, and Flux - Exercises 16.2 - Page 955: 15

#### Answer

$36$

#### Work Step by Step

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: $\int_C (x^2+y^2) \ dy=\int_{C_{1}} (x^2+y^2) \ dy+\int_{C_{2}} (x^2+y^2) \ dy =0+36=36$

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