1. **State the problem:** We need to evaluate the line integral of the scalar function $f(x,y,z) = x - 3y^2 + z$ over the curve $C = C_1 \cup C_2$, where: - $C_1$ is the line segment from $(0,0,0)$ to $(1,1,0)$ in the $xy$-plane. - $C_2$ is the line segment from $(1,1,0)$ to $(1,1,1)$ along the $z$-axis. 2. **Recall the line integral formula for scalar functions:** $$\int_C f(x,y,z) \, ds = \int_a^b f(\mathbf{r}(t)) \|\mathbf{r}'(t)\| \, dt$$ where $\mathbf{r}(t)$ parametrizes the curve $C$. 3. **Parametrize $C_1$:** Since $C_1$ goes from $(0,0,0)$ to $(1,1,0)$, a natural parametrization is: $$\mathbf{r}_1(t) = (t, t, 0), \quad t \in [0,1]$$ Then, $$\mathbf{r}_1'(t) = (1,1,0)$$ and $$\|\mathbf{r}_1'(t)\| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}$$ 4. **Evaluate $f$ along $C_1$:** $$f(\mathbf{r}_1(t)) = t - 3t^2 + 0 = t - 3t^2$$ 5. **Set up the integral over $C_1$:** $$\int_{C_1} f \, ds = \int_0^1 (t - 3t^2) \sqrt{2} \, dt = \sqrt{2} \int_0^1 (t - 3t^2) \, dt$$ 6. **Compute the integral over $C_1$:** $$\int_0^1 (t - 3t^2) \, dt = \left[ \frac{t^2}{2} - t^3 \right]_0^1 = \frac{1}{2} - 1 = -\frac{1}{2}$$ Therefore, $$\int_{C_1} f \, ds = \sqrt{2} \times \left(-\frac{1}{2}\right) = -\frac{\sqrt{2}}{2}$$ 7. **Parametrize $C_2$:** $C_2$ goes from $(1,1,0)$ to $(1,1,1)$ along the $z$-axis: $$\mathbf{r}_2(t) = (1, 1, t), \quad t \in [0,1]$$ Then, $$\mathbf{r}_2'(t) = (0,0,1)$$ and $$\|\mathbf{r}_2'(t)\| = 1$$ 8. **Evaluate $f$ along $C_2$:** $$f(\mathbf{r}_2(t)) = 1 - 3(1)^2 + t = 1 - 3 + t = -2 + t$$ 9. **Set up the integral over $C_2$:** $$\int_{C_2} f \, ds = \int_0^1 (-2 + t) \, dt$$ 10. **Compute the integral over $C_2$:** $$\int_0^1 (-2 + t) \, dt = \left[-2t + \frac{t^2}{2}\right]_0^1 = (-2 + \frac{1}{2}) - 0 = -\frac{3}{2}$$ 11. **Add the integrals over $C_1$ and $C_2$ to get the total integral:** $$\int_C f \, ds = \int_{C_1} f \, ds + \int_{C_2} f \, ds = -\frac{\sqrt{2}}{2} - \frac{3}{2}$$ **Final answer:** $$\boxed{\int_C f(x,y,z) \, ds = -\frac{\sqrt{2}}{2} - \frac{3}{2}}$$