I have a question that states: Calculate the work done by the vector field $$F = y^3 \hat{i} + 3xy^2 \hat{j}$$ A parametrized version is given as
$$x(t) = cos(t) +\frac{1}{4}sin(5t)^2$$ $$y(t) = sin(t)$$ $$t = 0... 2 \pi$$
I originally began by using Green's theorem,
Work = $\int F_xdx + F_ydy$
W = $\int Pdx + Qdy$
W = $\int y^3dx + 3xy^2dy$
Solving for $dx$ and $dy$ gave me...
$ dx = -sin(t) + \frac{5}{2}sin(5t)dt$
$dy= cos(t)dt$
I then attempted to input the equations with relation to $t$...
$\int_{0}^{2\pi} sin(t)^3(-sin(t) + 5/2sin(5t))dt + 3(cos(t) + 1/4sin(5t)^2)sin(t)^2cos(t)dt$
However, this integral seems extremely complex and I want to be sure I have the correct idea before proceeding. Is there another method that might be more effective to implement?
So did you try using Green's Theorem, which is in your title? What is $\displaystyle{\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}}$? (Note that the curve is a simple closed curve.)