## University Calculus: Early Transcendentals (3rd Edition)

$1$
Since, we have $ds=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$ or, $ds=\sqrt{(1)^2+( 3 )^2+(1)^2} dt \implies ds= \sqrt {3} dt$ Now, the line integral is: $\int_C \dfrac{\sqrt 3}{x^2+y^2+z^2} ds=\int_{1}^{\infty} \dfrac{\sqrt 3}{t^2+t^2+t^2} (\sqrt{3}) dt$ or, $=\int_{1}^{\infty} \dfrac{3}{3t^2} dt$ or, $=[\dfrac{-1}{t}]_{1}^{\infty}$ or, $=1$