How to calculate the line integral of a vector field over a parabola I am trying to answer a question about line integrals, I have had a go at it but I am not sure where I am supposed to incorporate the line integral into my solution.
$$ \mathbf{V} = xy\hat{\mathbf{x}} + -xy^2\hat{\mathbf{y}}$$
$$ \mathrm{d}\mathbf{l} = \hat{\mathbf{x}}\mathrm{d}x + \hat{\mathbf{y}}\mathrm{d}y $$
$$ \int_C\!\mathbf{V}\cdot\mathrm{d}\mathbf{l} = \int\!xy\,\mathrm{d}x - \int\!xy^2\,\mathrm{d}y = \left[ \frac{x^2y}{2}\right ]_?^? - \left[ \frac{xy^3}{3} \right]_?^? $$
I have a feeling that the parabola in question must come into play in the limits of the integrals, although I dont know how they are supposed to. The parabola in question is $y = \frac{x^2}{3}$ and the coordinates at which the line integral is supposed to go over are $a=(0,0)$ and $b=(3,3)$.
 A: *

*One could also use a parametrization.

Set $x(t)=t$ and and $y(t)= t^2/3$.

*

*The parabola corresponds to the endpoint of the position vector :

$ \vec r (t)= t\vec i + \frac {t^2}{3} \vec j$.

*

*This allows to calculate the differential $d\vec r$:

$d\vec{\mathbf{r}}(t)=\vec{\mathbf{r\space '}}(t) dt = (1\vec i + (2t/3) \vec j) dt$

*

*With the parametrization chosen, we get $\vec{\mathbf{V}}(t)= (t. \frac{t^2}{3})\vec i - (t. (\frac{t^2}{3})^2)\vec j$


*One can then apply  the definition : vector  line integral = $\int_C \vec V(t).d\vec r(t)= \int_{t_1}^{t_2} \vec (V(t).\vec r\space '(t) )dt$ ,
with, here,  $t_1 = 0$ and $t_2 =3$.
A: In this case, it's as simple as plugging in $y=\frac{x^2}{3}$ into the first integral and integrating from $x=0$ to $x=3$. In the second integral, do the same thing, but plug in $dy=\frac{2x}{3}dx$. In other words, parameterize the whole curve in terms of $x$.
In other examples (like circles), it may be more convenient to parameterize $x$ and $y$ with a different variable.
