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.

 

[paranormal]

Your attempt at a solution is correct. However, when you plug in 0 for \varphi, you are only looking at the work done in the r direction. To find the total work done, you need to integrate over both r and \varphi.

The integral for the work done in the r direction is:
\int_0^2 r^3\cos^2\varphi\sin\varphi dr

To find the total work done, you also need to integrate over \varphi. The integral for the work done in the \varphi direction is:
\int_0^{\pi/2} r^4\cos\varphi\cos(2\varphi) d\varphi

Therefore, the total work done is:
\int_0^2\int_0^{\pi/2} r^3\cos^2\varphi\sin\varphi + r^4\cos\varphi\cos(2\varphi) dr d\varphi

You can solve this integral to find the total work done.