Use Green's theorem to evaluate $\int_{C} F \cdot d r$. (Check the orientation of the curve before applying the theorem.) Risk of Rain 2 Sid Meier's CivilizationV $F(x, y)=\left\langle\sqrt{x^{2}+2}, \tan ^{-1}(x)\right\rangle, C$ is the triangle from $(0,0)$ to $(1,1)$ to $(0,1)$ to $(0,0)$

Step 1 ：First, we need to compute the curl of the vector field $F$. The curl of a vector field $F = \langle P, Q \rangle$ in two dimensions is given by $\nabla \times F = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$.

Step 2 ：Here, $P = \sqrt{x^2 + 2}$ and $Q = \tan^{-1}(x)$. So, we need to compute $\frac{\partial Q}{\partial x}$ and $\frac{\partial P}{\partial y}$.

Step 3 ：We have $\frac{\partial Q}{\partial x} = \frac{1}{1+x^2}$ and $\frac{\partial P}{\partial y} = 0$ (since $P$ does not depend on $y$).

Step 4 ：Therefore, $\nabla \times F = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{1}{1+x^2} - 0 = \frac{1}{1+x^2}$.

Step 5 ：Now, we apply Green's theorem, which states that $\int_C F \cdot dr = \int\int_D (\nabla \times F) \, dA$, where $D$ is the region enclosed by the curve $C$.

Step 6 ：In this case, $D$ is the triangle with vertices at $(0,0)$, $(1,1)$, and $(0,1)$. We can integrate over this region by letting $x$ go from $0$ to $1$ and $y$ go from $0$ to $1-x$.

Step 7 ：So, we have $\int_C F \cdot dr = \int_0^1 \int_0^{1-x} \frac{1}{1+x^2} \, dy \, dx$.

Step 8 ：This simplifies to $\int_0^1 \frac{1-x}{1+x^2} \, dx$.

Step 9 ：This integral can be computed using a standard calculus technique. The result is $\frac{1}{2} \ln(2) - 1$.

Step 10 ：Therefore, the value of the line integral $\int_C F \cdot dr$ is $\boxed{\frac{1}{2} \ln(2) - 1}$.