1. **Problem statement:** We want to set up the integral $$\int_C 6x \, ds$$ where curve $C$ consists of three parts: - Part 1: $y = 3 + x^2$ from $x = -2$ to $x = 0$ - Part 2: $y = 3 - x^2$ from $x = 0$ to $x = 2$ - Part 3: line segment from $(2,-1)$ to $(-1,-2)$ 2. **Formula for line integrals with respect to arc length:** $$\int_C f(x,y) \, ds = \int_a^b f(x(t), y(t)) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$$ where $t$ parametrizes the curve. 3. **Part 1: Curve $y = 3 + x^2$, $x \in [-2,0]$** - Parametrize by $x = t$, $y = 3 + t^2$, $t \in [-2,0]$ - Compute derivatives: $$\frac{dx}{dt} = 1, \quad \frac{dy}{dt} = 2t$$ - Compute $ds$: $$ds = \sqrt{1^2 + (2t)^2} \, dt = \sqrt{1 + 4t^2} \, dt$$ - The integrand is $6x = 6t$ - Integral for part 1: $$\int_{-2}^0 6t \sqrt{1 + 4t^2} \, dt$$ 4. **Part 2: Curve $y = 3 - x^2$, $x \in [0,2]$** - Parametrize by $x = t$, $y = 3 - t^2$, $t \in [0,2]$ - Compute derivatives: $$\frac{dx}{dt} = 1, \quad \frac{dy}{dt} = -2t$$ - Compute $ds$: $$ds = \sqrt{1^2 + (-2t)^2} \, dt = \sqrt{1 + 4t^2} \, dt$$ - The integrand is $6x = 6t$ - Integral for part 2: $$\int_0^2 6t \sqrt{1 + 4t^2} \, dt$$ 5. **Part 3: Line segment from $(2,-1)$ to $(-1,-2)$** - Parametrize line segment: Let $\mathbf{r}(s) = (x(s), y(s)) = (2, -1) + s((-1) - 2, (-2) - (-1)) = (2 - 3s, -1 - s)$ - Parameter $s \in [0,1]$ - Compute derivatives: $$\frac{dx}{ds} = -3, \quad \frac{dy}{ds} = -1$$ - Compute $ds$: $$ds = \sqrt{(-3)^2 + (-1)^2} \, ds = \sqrt{9 + 1} \, ds = \sqrt{10} \, ds$$ - The integrand $6x = 6(2 - 3s) = 12 - 18s$ - Integral for part 3: $$\int_0^1 (12 - 18s) \sqrt{10} \, ds$$ 6. **Final integral setup:** $$\int_C 6x \, ds = \int_{-2}^0 6t \sqrt{1 + 4t^2} \, dt + \int_0^2 6t \sqrt{1 + 4t^2} \, dt + \int_0^1 (12 - 18s) \sqrt{10} \, ds$$ This completes the setup of the integral over the curve $C$ as requested.