1. **Problem statement:** Calculate the line integral of the function $f(x,y,z) = x + \sqrt{y - z^2}$ along two curves from $(0,0,0)$ to $(1,1,1)$. 2. **Formula:** The line integral for scalar function $f$ along curve $\mathbf{r}(t)$ 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 magnitude of the derivative of $\mathbf{r}(t)$. --- ### Curve $C_1$: $\mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{j} + 0\mathbf{k}$, $0 \leq t \leq 1$ 3. Parametrize: $x = t$, $y = t^2$, $z = 0$ 4. Compute $f(t)$: $$f(t) = t + \sqrt{t^2 - 0^2} = t + t = 2t$$ 5. Compute $\mathbf{r}'(t)$: $$\mathbf{r}'(t) = (1, 2t, 0)$$ 6. Magnitude: $$|\mathbf{r}'(t)| = \sqrt{1^2 + (2t)^2 + 0^2} = \sqrt{1 + 4t^2}$$ 7. Integral setup: $$I_1 = \int_0^1 2t \sqrt{1 + 4t^2} \, dt$$ 8. Substitution: Let $u = 1 + 4t^2$, then $du = 8t \, dt$, so $t \, dt = \frac{du}{8}$. 9. Rewrite integral: $$I_1 = \int_1^5 2t \sqrt{u} \, dt = \int_1^5 \sqrt{u} \frac{du}{4} = \frac{1}{4} \int_1^5 u^{1/2} \, du$$ 10. Integrate: $$\frac{1}{4} \cdot \frac{2}{3} u^{3/2} \Big|_1^5 = \frac{1}{6} (5^{3/2} - 1)$$ --- ### Curve $C_2$: $\mathbf{r}(t) = \mathbf{i} + \mathbf{j} + t\mathbf{k}$, $0 \leq t \leq 1$ 11. Parametrize: $x=1$, $y=1$, $z=t$ 12. Compute $f(t)$: $$f(t) = 1 + \sqrt{1 - t^2}$$ 13. Compute $\mathbf{r}'(t)$: $$\mathbf{r}'(t) = (0,0,1)$$ 14. Magnitude: $$|\mathbf{r}'(t)| = 1$$ 15. Integral setup: $$I_2 = \int_0^1 \left(1 + \sqrt{1 - t^2}\right) dt = \int_0^1 1 \, dt + \int_0^1 \sqrt{1 - t^2} \, dt$$ 16. Integrate parts: $$\int_0^1 1 \, dt = 1$$ $$\int_0^1 \sqrt{1 - t^2} \, dt = \frac{\pi}{4}$$ (area of quarter circle radius 1) 17. Final integral: $$I_2 = 1 + \frac{\pi}{4}$$ --- **Final answers:** $$\boxed{\int_{C_1} f \, ds = \frac{1}{6} (5^{3/2} - 1)}$$ $$\boxed{\int_{C_2} f \, ds = 1 + \frac{\pi}{4}}$$ This is a clear step-by-step solution suitable for school homework.