1.Evaluate the differential form $\displaystyle ydx + zdy + xdz $ on the following line segment: $\displaystyle (0,2,-1) $ to $\displaystyle (3,0,1) $.

So $\displaystyle \begin{cases} x = 3t \\ y = 2-2t \ \ \ \ \ \ 0 \leq t \leq 1 \\ z = 2t-1 \end{cases} $ so that $\displaystyle dx = 3dt, \ \ dy = -2dt, \ \ dz = 2dt $.

Then $\displaystyle \int_{(0,2,-1)}^{(3,0,1)} ydx + zdy + xdz = \int_{0}^{1} (2-2t)(3dt) + (2t-1)(-2dt) + (3t)(2dt) $

Is this correct?