Given points in $\mathbb R^3$: $$A=(1,0,0)\;\;,\;\; B=(1,1,0)\;\;,\;\; C=(1,1,1)$$
Then compute $$\int_{C}^{ }ydx+zdy+xdz$$
- Along the line $OC$
- Along $OABC$
Parametric equation of the first line is given by $$x=t$$ $$y=t$$ $$z=t$$
Where $ t \in [0,1]$,so: $$\int_{\overline{OC}}^{ }ydx+zdy+xdz=3\int_{0}^{1 }tdt=3/2$$
The equation of the lines $OA,AB,BC$ is given by (resp) $$x=t\;\;,\;\;y=0\;\;,\;\;z=0$$$$x=-t^2+2t\;\;,\;\;y=t\;\;,\;\;z=0$$$$x=-t^2+2t\;\;,\;\;y=-t^2+2t\;\;,\;\;z=t$$
So: $$\int_{OABC}^{ }ydx+zdy+xdz$$$$=\int_{\overline{OA}}^{ }ydx+zdy+xdz+\int_{\overline{AB}}^{ }ydx+zdy+xdz+\int_{\overline{BC}}^{ }ydx+zdy+xdz$$ $$=0+\int_{0}^{1}t\left(-2t+2\right)dt+\int_{0}^{1}\left[\left(-t^{2}+2t\right)\left(-2t+2\right)+t\left(-2t+2\right)+\left(-t^{2}+2\right)\right]dt$$$$=\int_{0}^{1}\left(2t^{3}-9t^{2}+6t+2\right)dt=\frac{5}{2}$$
I want to know how much of my work is correct.
Your answer for $OC$ is correct but not for $OABC$.
Your parametrization for $OA$ is correct and the line integral over line segment $OA$ is zero as you mentioned.
Line segment $AB$: $x=-t^2+2t\;\;,\;\;y=t\;\;,\;\;z=0, \ 0 \leq t \leq 1$ is not correct parametrization. $t=0$ does not give you point $A (1, 0, 0)$.
Rather use parametrization,
$(1, 0, 0) + [(1, 1, 0) - (1, 0, 0)] t = (1, t, 0), 0 \leq t \leq 1$
So line integral is $\displaystyle \int_0^1 (t, 0, 1) \cdot (0, 1, 0) \ dt = 0$
Similarly, parametrization of line segment $BC$:
$r(t) = (1, 1, t), 0 \leq t \leq 1$
line integral is $\displaystyle \int_0^1 (1, t, 1) \cdot (0, 0, 1) \ dt = 1$
So over path $OABC$, the line integral is $1$.