1. **Problem statement:** Calculate the line integral of the function $f(x,y,z) = x + \sqrt{y - z^2}$ along two given curves from $(0,0,0)$ to $(1,1,1)$. 2. **Recall the line integral formula:** For a scalar function $f$ and a curve $\mathbf{r}(t)$, the line integral is $$\int_C f(x,y,z)\, ds = \int_a^b f(\mathbf{r}(t)) \|\mathbf{r}'(t)\| dt$$ where $\mathbf{r}'(t)$ is the derivative of the position vector. --- ### Curve $C_1$: $\mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{j} + 0\mathbf{k}$, $0 \leq t \leq 1$ 3. Parametrize $x,y,z$: $$x = t, \quad y = t^2, \quad z = 0$$ 4. Compute $f$ along $C_1$: $$f(t) = t + \sqrt{t^2 - 0^2} = t + t = 2t$$ 5. Compute $\mathbf{r}'(t)$: $$\mathbf{r}'(t) = \frac{d}{dt}(t, t^2, 0) = (1, 2t, 0)$$ 6. Compute the magnitude: $$\|\mathbf{r}'(t)\| = \sqrt{1^2 + (2t)^2 + 0^2} = \sqrt{1 + 4t^2}$$ 7. Set up the integral: $$I_1 = \int_0^1 2t \sqrt{1 + 4t^2} \, dt$$ 8. Use substitution: Let $u = 1 + 4t^2$, then $du = 8t dt$, so $t dt = \frac{du}{8}$. Rewrite the integral: $$I_1 = \int_{u=1}^{5} 2t \sqrt{u} \, dt = \int_1^5 2t \sqrt{u} \, dt$$ Since $t dt = \frac{du}{8}$, then $2t dt = \frac{du}{4}$. So, $$I_1 = \int_1^5 \sqrt{u} \frac{du}{4} = \frac{1}{4} \int_1^5 u^{1/2} du$$ 9. Integrate: $$\frac{1}{4} \cdot \frac{2}{3} u^{3/2} \Big|_1^5 = \frac{1}{6} (5^{3/2} - 1)$$ --- ### Curve $C_2$: $\mathbf{r}(t) = \mathbf{i} + \mathbf{j} + t\mathbf{k}$, $0 \leq t \leq 1$ 10. Parametrize $x,y,z$: $$x = 1, \quad y = 1, \quad z = t$$ 11. Compute $f$ along $C_2$: $$f(t) = 1 + \sqrt{1 - t^2}$$ 12. Compute $\mathbf{r}'(t)$: $$\mathbf{r}'(t) = (0,0,1)$$ 13. Magnitude: $$\|\mathbf{r}'(t)\| = 1$$ 14. Set up the integral: $$I_2 = \int_0^1 \left(1 + \sqrt{1 - t^2}\right) dt = \int_0^1 1 dt + \int_0^1 \sqrt{1 - t^2} dt$$ 15. Integrate the first part: $$\int_0^1 1 dt = 1$$ 16. Integrate the second part (area of a quarter circle of radius 1): $$\int_0^1 \sqrt{1 - t^2} dt = \frac{\pi}{4}$$ 17. So, $$I_2 = 1 + \frac{\pi}{4}$$ --- ### **Final answers:** $$\boxed{\int_{C_1} f \, ds = \frac{1}{6} (5^{3/2} - 1)}$$ $$\boxed{\int_{C_2} f \, ds = 1 + \frac{\pi}{4}}$$