Finding a Line Integral Along \(C\) with Respect to \(x\) and with Respect to \(y\)

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 \[ \begin{eqnarray*} \int_{C}(x-3y)\,dx &=&\int_{1}^{2}(t^{2}-3t)2t\,dt=2\int_{1}^{2}(t^{3}-3t^{2}) dt=-\dfrac{13}{2} \notag \\[4pt] \int_{C}(x-3y)\,dy &=&\int_{1}^{2}(t^{2}-3t)\,dt=-\dfrac{13}{6} \end{eqnarray*} \]