Let $\mathbf{F}(\mathbf{r})=(x^{2}z,xy^{2},z^{2})$ and define a closed curve $C$ in $\Bbb{R}^{3}$ comprised of three straight line segments $C_{1},C_{2}$ and $C_{3}$ formed by the intersection of the plane $x+y+z=1$ with the planes $y=0,z=0$ and $x=0$ (respectively).
Sketch the curve $C=C_{1}\cup C_{2}\cup C_{3}$ in the $(x,y,z)$-space, and calculate $\displaystyle\int_{C}\mathbf{F}\cdot d\mathbf{r}$ by considering each line segment separately. Be careful to make sure the orientation of each segment of $C$ is aligned with the other segments.
Below is an image that I have plotted for all four planes: orange represents $x+y+z=1$, yellow represents $y=0$, blue represents $x=0$, and red represents $z=0$. I am not too sure how to evaluate the line integral, and how to relate the curve $C$ back to $\mathbf{F}(\mathbf{r})$ - appreciate any pointers I can get, thanks.
Intersection points of $x + y + z = 1$ plane with $x = 0, y = 0, z = 0$ planes are,
$A (1,0,0), B (0, 1, 0), C (0, 0, 1)$.
So our lines (all oriented in the same direction) are
$AB = (1, 0, 0) + (-1, 1, 0)t \, $ ($t = 0$ to $t = 1$)
$BC = (0, 1, 0) + (0, -1, 1)t \, $ ($t = 0$ to $t = 1$)
$CA = (0, 0, 1) + (1, 0, -1)t \, $ ($t = 0$ to $t = 1$)
$i)$ Now for our line integral along path $AB$
Please note $r(t) = (1-t, t, 0) \,$ and $\, r'(t) = (-1,1,0)$
$C_1 = \displaystyle \int_{t=0}^{1} (x^{2}z,xy^{2},z^{2}) dr = \int_{t=0}^{1} ((1-t)^2.0,(1-t)t^2, 0) r'(t) dt$
$C_1 = \displaystyle \int_{t=0}^{1} (0,(t^2-t^3), 0) \cdot (-1,1,0) dt = \int_{t=0}^1 (t^2-t^3) dt = \frac{1}{12}$
You can similarly find $C_2$ and $C_3$.