Answer
$94.8231$
Work Step by Step
Here, $ds=\sqrt{(dx)^2+(dy)^2+(dz)^2}$
or, $ds=\sqrt{(dx/dt)^2+(dy/dt)^2+(dz/dt)^2}$
Thus, $ds=\sqrt{(2t)^2+(3t^2)^2+(1/2\sqrt t)^2}=\sqrt {4t^2+9t^4+1/4t} dt$
Now, $\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=\int_1^2 (t^2)(t^3) \tan^{-1} \sqrt t \sin (t^3+t^4) \sqrt {4t^2+9t^4+1/4t} dt$
By using the calculator, we have $\int_{C} \overrightarrow{F} \cdot \overrightarrow{dr}=94.8231$
