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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.

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  • $\begingroup$ $d\vec{r} = (3t^2 dt, 2t dt)$. Compute the dot product like you would for any two vectors. $\endgroup$
    – Charles Hudgins
    Commented Feb 15, 2022 at 22:21
  • $\begingroup$ @CharlesHudgins Ahh, thank you. I was using $\|r'(t)\|=\sqrt{\frac{dx}{dt}^2+\frac{dy}{dt}^2}$ as my $d\vec{r}$. $\endgroup$
    – brandon_ducks
    Commented Feb 15, 2022 at 22:29

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