Find $$\int_{C}(x-3y)\,dx\qquad\hbox{and}\qquad \int_{C}(x-3y)\,dy$$

if $C$ is the part of the parabola $x=y^{2}$ that joins the points $(1,1)$ and $(4,2)$.

Solution The function $f(x,y) =x-3y$ is continuous, and $C$ is a smooth curve everywhere in the plane. So, we can use the formulas (3) and (4). Using the parametric equations of the curve $C,$ $x=t^{2}$ and $y=t$, $\ 1\leq t\leq 2$, we find $dx=2t\,dt$ and $dy=dt$. Then $$\begin{eqnarray*} \int_{C}(x-3y)\,dx &=&\int_{1}^{2}(t^{2}-3t)2t\,dt=2\int_{1}^{2}(t^{3}-3t^{2}) dt=-\dfrac{13}{2} \notag \\[4pt] \int_{C}(x-3y)\,dy &=&\int_{1}^{2}(t^{2}-3t)\,dt=-\dfrac{13}{6} \end{eqnarray*}$$