Find the line integral $\int_{C}(y^{2}\,dx-x^{2}\,dy)$ along

$C_{1}$: The parabola $y=x^{2}$ joining the two points $(0,0)$ and $(2,4)$

$C_{2}$: The line $y=2x$ joining the two points $(0,0)$ and $(2,4)$

Solution The curves $C_{1}$ and $C_{2}$ are smooth, and the functions $P(x,y) =y^{2}$ and $Q(x,y) = -x^{2}$ are continuous in the $xy$-plane.

Along $C_{1},$ parametric equations for the parabola are $x(t) =t,$ $y(t) =t^{2}.$ Then $dx=dt$ and $dy=2t\,dt,$ so $$\begin{eqnarray*} \int_{C_{1}}(y^{2}\,dx-x^{2}\,dy)&=&\int_{C_{1}}y^{2}\,dx-\int_{C_{1}}x^{2}\,dy=\int_{0}^{2}(t^2) ^{2}\,dt-\int_{0}^{2}t^{2}(2t\,dt)\\ &=& \int_{0}^{2}t^{4}\,dt-2\int_{0}^{2}t^{3}\,dt=\left[ \dfrac{t^{5}}{5}\right] _{0}^{2}-\left[ \dfrac{t^{4}}{2}\right] _{0}^{2}=-\dfrac{8}{5} \end{eqnarray*}$$

Figure 18

Along $C_{2}$, parametric equations for the line segment are $x(t) =t$, $y(t) =2t.$ Then $dx=dt$ and $dy=2\,dt,$ so $$\begin{eqnarray*} \int_{C_{2}}(y^{2}\,dx-x^{2}\,dy)&=&\int_{C_{2}}y^{2}\,dx-\int_{C_{2}}x^{2}\,dy=\int_{0}^{2}(2t) ^{2}\,dt-\int_{0}^{2}t^{2}(2\,dt)\\[4pt] &=& \left[ \dfrac{4t^{3}}{3}\right] _{0}^{2}-\left[ \dfrac{2t^{3}}{3}\right] _{0}^{2}=\dfrac{16}{3} \end{eqnarray*}$$

Figure 18 shows the curves $C_{1}$ and $C_{2}$ joining the points $( 0,0)$ and $(2,4) .$