Find $\int_{C}\mathbf{F}\,{\boldsymbol\cdot}\, d\mathbf{r}$ if $$\begin{equation*} \mathbf{F}(x,y,z)=xy^{2}\,\mathbf{i}+x^{2}z\,\mathbf{j}-(y-x)\,\mathbf{k} \end{equation*}$$

and the curve $C$ is traced out by $\mathbf{r}(t)=t\,\mathbf{i}+t^{2}\mathbf{j}+t^{3}\,\mathbf{k}$, $\ 0\leq t\leq 1$.

Solution Parametric equations of the curve $C$ are $$\begin{equation*} x=t\qquad y=t^{2}\qquad z=t^{3}\qquad 0 \le t \le 1 \end{equation*}$$

So, $$\mathbf{F}=xy^{2}\,\mathbf{i}+x^{2}z\,\mathbf{j}-(y-x)\,\mathbf{k}=t^{5} \mathbf{i}+t^{5}\mathbf{j}-(t^{2}-t)\,\mathbf{k}$$

and $$d\mathbf{r}=\dfrac{d\mathbf{r}}{dt}\,dt=(\mathbf{i}+2t\mathbf{j}+3t^{2} \mathbf{k})\,dt$$

Then $$\begin{eqnarray*} \mathbf{F}\,{\cdot}\, d\mathbf{r}&=&[ t^{5}\mathbf{i}+t^{5}\,\mathbf{j} -(t^{2}-t)\,\mathbf{k}] \,{\cdot}\, ( \mathbf{i}+2t\,\mathbf{j}+3t^{2}\, \mathbf{k}) \,dt=[ t^{5}+2t^{6}-3t^{2}(t^{2}-t)] \,dt\\ &=&( 2t^{6}+t^{5}-3t^{4}+3t^{3}) \,dt \end{eqnarray*}$$

so that $$\int_{C}\mathbf{F}\,{\boldsymbol\cdot}\, d\mathbf{r}=\int_{0}^{1}(2t^{6}+t^{5}-3t^{4}+3t^{3}) \,dt=\dfrac{253}{420}$$