## Calculus: Early Transcendentals 8th Edition

$1.72599$
Here, $ds=\sqrt{(dx)^2+(dy)^2+(dz)^2}$ or, $ds=\sqrt{(dx/dt)^2+(dy/dt)^2+(dz/dt)^2}$ Thus, $ds=\sqrt{(3)^2+(2t)^2+(4t^3)^2}dt=\sqrt {9+4t^2+16t^{16}} dt$ Now, $\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_{-1}^1 t^4 \ln [(1+3t)+(2+t^2)] (\sqrt {9+4t^2+16t^{16}} dt)$ By using the calculator, we have $\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=1.72599$