I have a vector field defined as $\vec{F} = xy \hat{\textbf{i}} - y^2 \hat{\textbf{j}}$ and am travelling through this field along a curve defined as $C = \frac{1}{4}x^2$. To calculate the work done as I travel through the vector field between (0,0) and (2,1), I am parametising with $t$ and then evaluating the line integral in terms of this parameter. For some reason when I parametise with $x = 2t$ I get a different answer to when I parametise with $x = t$ even though I change my bounds to compensate for this. Below are the two attempts with different results for work done:
with $x = 2t$:
$x = 2t$ and $y = \frac{1}{4}x^2$ hence $y = t^2$.
Differentiating $x$ and $y$ with respect to $t$ shows $dx = 2dt$ and $dy = 2t dt$
Substituting this into the equation for work done and setting bounds as $1$ and $0$ (so that we start at $x=0, y=0$ for $t=0$ and end at $x=2, y=1$ for $t=1$:
$\int^{(2,0)}_{(0,0)} \vec{F} \cdot d\vec{r} = \int^{(2,0)}_{(0,0)} (xy dx - y^2 dy)\cdot d\vec{r} = \int^{1}_{0} (4t^3 - 2t^5)dt = \frac{2}{3}$
with $\vec{r} = x \hat{\textbf{i}} + y \hat{\textbf{j}}$
with $x = t$:
$x = t$ and $y = \frac{1}{4}x^2$ hence $y = \frac{1}{4} t^2$.
Differentiating $x$ and $y$ with respect to $t$ shows $dx = dt$ and $dy = \frac{1}{2}t dt$
Substituting this into the equation for work done and setting bounds as $1$ and $0$ (so that we start at $x=0, y=0$ for $t=0$ and end at $x=2, y=1$ for $t=0$:
$\int^{(2,0)}_{(0,0)} \vec{F} \cdot d\vec{r} = \int^{(2,0)}_{(0,0)} (xy dx - y^2 dy)\cdot d\vec{r} = \frac{1}{4}\int^{2}_{0} (t^3 - t^2)dt = \frac{1}{3}$
with $\vec{r} = x \hat{\textbf{i}} + y \hat{\textbf{j}}$
In your second integral, $y^2\,dy=\frac{1}{16}t^4\cdot\frac{1}{2}t\,dt=\frac{1}{32}t^5\,dt$. You should then get:
$$\oint_C\vec F\cdot\,d\vec r=\frac{1}{4}\int_0^2(t^3-\frac{1}{8}t^5)\,dt=\frac{1}{16}\cdot2^4-\frac{1}{192}\cdot2^6=1-\frac{1}{3}=\frac{2}{3}$$
And all is well.