Twisted minds

Each component of $F$ is given in terms of the x, y, and z components of $\textbf{r}(t)$ . i.e., the x-component of $F$ is the product of the x and y components of $\textbf{r}(t)$ , the y-component of $F$ is the product of the y and z components of $\textbf{r}(t)$ , and the z-component of $F$ is the product of the z and x components of $\textbf{r}(t)$ . So our line integral is: $\begin{aligned} N=\int_C \! F\cdot \mathrm{d}r &= \int_C \! F(\textbf{r}(t))\cdot\textbf{r}'\!(t) \ \mathrm{d}t \\ & =\int_C \! ((t\cdot t^2), (t^2 \cdot t^3), (t^3 \cdot t))\cdot(1, 2t, 3t^2) \ \mathrm{d}t \\ &=\int_C \! (t^3, t^5, t^4)\cdot(1, 2t, 3t^2) \ \mathrm{d}t \end{aligned}$ Taking the dot product and integrating from 0 to 1: $\Rightarrow \int_{0}^{1} \! (t^3+5t^6) \ \mathrm{d}t = \left[\frac{1}{4}t^4+\frac{5}{7}t^7\right]_{0}^{1} = \frac{27}{28} \approx \boxed{0.9643}$