Dance of the mosquitos

$$ From function $\mathbf{r}(t)=11t^4\;\mathbf{i}+t^3\;\mathbf{j}$ , we have $\begin{aligned} x&=11t^4 \;\;\;\;\Rightarrow\;\;\; dx=44t^3\,dt,\\ y&=t^3 \;\;\;\;\;\;\;\,\Rightarrow\;\;\; dy=3t^2\,dt.\\ \end{aligned}$ Thus, $\begin{aligned} \int_C \mathbf{F}\cdot d\mathbf{r}&=\int_C (xy\,dx+3y^2\,dy)\\ &=\int_0^1 (484t^{10}+9t^8)\,dt\\ &=\left[44t^{11}+t^9\right]_0^1\\ &=\boxed{45} \end{aligned}$ $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$

Well, first observe that $\nabla \times \mathbf{F} \neq 0$ , so there is no scalar field $G$ such that $\nabla G = \mathbf{F}$ . In other words, we have to compute the integral by its definition. So:

$\int_{C} \mathbf{F} \cdot d\mathbf{r} = \int_{0}^{1} \mathbf{F(r(\mathit{t}))} \cdot \mathbf{r'(\mathit{t})} dt$

We also have $\mathbf{r'(\mathit{t})} = 44t^{3} \mathbf{i} + 3t^{2}\mathbf{j}$ and $\mathbf{F(r'(\mathit{t}))} = 11t^{7}\mathbf{i} + 3t^{6}\mathbf{j}$ which implies that $\mathbf{F(r(\mathit{t}))} \cdot \mathbf{r'(\mathit{t})} = 484t^{10} + 9t^{8}$ . Therefore the integral becomes

$\int_{0}^{1} \mathbf{F(r(\mathit{t}))} \cdot \mathbf{r'(\mathit{t})} = \int_{0}^{1} 484t^{10} + 9t^{8} dt = 45$