I'm trying to find the work done by the force $\vec{F}=(x^2,2xy)$ along the path that can be given parametrically by $x=t^3$ and $y=t^2$. The path starts at the origin and ends at $(1,1)$
I believe my force can now be represented by: $\vec{F}=(t^6,2t^5)$
With my $\vec{r}$ being $\vec{r}=(t^3, t^2)$
This leads my to believe that the work can now be represented by the integral:
$$W = \int \vec{F} \cdot d\vec{r} = \int_0^1 (t^6, 2t^5) \cdot \sqrt{9t^4+4t^2}dt$$
Which would, I believe, simplify to
$$W = \int_0^1 t^6 \sqrt{9t^4+t^2} + 2t^5 \sqrt{9t^4 + 4t^2}dt $$
The answer to this indefinite integral is reasonable, however the definite integral is nasty looking. This class can be done without a calculator, which is why I think my steps to setting up are incorrect.