Finding work done in spherical coordinates

[HeisenbergW]

1. Find the work done by the force F=r³*cos²\varphi*sin\varphi*\hat{r} + r³*cos\varphi*cos(2\varphi) \hat{\varphi}
from the point (0,0,0) to (2,0,0)

Homework Equations

Work=\int F*dr
where dr= dr\hat{r} + rd\varphi\hat{\varphi}

The Attempt at a Solution

When muliplying the line element, dr, by the force, F, I come up with
\int r³*cos²\varphi*sin\varphi dr +\int r⁴*cos\varphi*cos(2\varphi) d\varphi

I believe the r goes from 0 to 2, and there is no change in \varphi

I end up with 4*cos^{2}\varphi*sin\varphi
but then when I plug in 0 for \varphi, the answer ends up being zero, which I have a hard time believing since it moves from 0 to 2.
Any help is greatly appreciated
Thank You in advance.

 

[Simon Bridge]

F=r³*cos²\varphi*sin\varphi*\hat{r} + r³*cos\varphi*cos(2\varphi) \hat{\varphi}

What is the force along \varphi = 0?
(This should simplify your line integral.)

 

[vb2341]

Check the force along the path you are given (because it's really a line integral through a vector field), and it should become fairly simple to see why that is. Notice that your psi component didn't really change in the integral that mattered.

 

[HeisenbergW]

Thanks for the replies
I believe you are saying that the force along \varphi=0 is just zero along the r component, which is the only component that matters, since there is no motion in the other two coordinates. Since my force is actually zero at \varphi=0, it doesn't matter that I went from (0,0,0) to (2,0,0), since no force means no work.

Am I interpreting your comments correctly?
Thanks for the feedback. Always appreciated.