Chapter 15 - Section 15.1 - Line Integrals - Exercises - Page 827: 18

$-4a^2$

Here, we have $ ds=\sqrt {(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$ and $ds= a dt$ This implies that $L=\int_{0}^{2 \pi} -\sqrt {(0)^2+a^2 \sin^2 t}( a dt)$ or, $2a^2 \int_{0}^{\pi} -\sin t dt=2a^2(-1-1)\sqrt 3 [\ln (b)-\ln (a)]$ Thus, the line integral $L=-4a^2$