How Is Work Calculated Along a Semicircle in Vector Calculus?

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Homework Statement


Find the work done along the semicircle (x-1)2+y2=1 from the point (0,0,0) to (2,0,0)
When F=r3cos2φsinφ[itex]\hat{r}[/itex] + r3*cosφ*cos(2φ) [itex]\hat{φ}[/itex]

Homework Equations


Work
W=∫F*dr
where dr=dr= dr[itex]\hat{r}[/itex] + rsinφdφ[itex]\hat{φ}[/itex]

The Attempt at a Solution



I convert the equation of the line (x-1)2+y2=1 to cylindrical coordinates.
Where x=rsinφ
and y=rcosφ
so (x-1)2+y2=1
eventually becomes r=2cosφ

now i plug in this r value into the force vector given, which gives me F=8cos5φsinφ[itex]\hat{r}[/itex] + 8*cos4φ*cos(2φ) [itex]\hat{φ}[/itex]

W=∫F*dr

So now i dot product this with the dr, which is now dr= dr[itex]\hat{r}[/itex] + 2cosφ
sinφdφ[itex]\hat{φ}[/itex] due to the r value I found.

The dot product comes out to be

∫8cos5φsinφdr + 16*cos5φ*cos(2φ)sinφdφ

and because r=2cosφ, dr=-2sinφ dφ

so this is once again plugged into the integral of the dot product, creating

∫-16cos5φsin2φdφ + 16*cos5φ*cos(2φ)sinφdφ

where I integrate from pi/2 to 0, since it ends at x=2 and y=0, but begins at x=0 and y=0

This gives me -4/35, which does not seem to be a correct answer
Any mistakes?
Any help is appreciated
Thank You in advance
 
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HeisenbergW said:
where dr=dr= dr[itex]\hat{r}[/itex] + rsinφdφ[itex]\hat{φ}[/itex]

The equation is not correct.The second terms has to be rdφ[itex]\hat{φ}[/itex].
And make the vector [itex]\vec{dr}[/itex] different from dr, change of the coordinate r.


ehild
 
For d[itex]\vec{r}[/itex] I actually left out the second term, which is the elevation angle, since it is constant. I guess I should have clarified that I am using this in spherical coordinates, but I did produce a typo It should be
dr[itex]\hat{r}[/itex]+rsinθd[itex]\varphi[/itex][itex]\hat{\varphi}[/itex].
Which of course just simplifies to
dr[itex]\hat{r}[/itex]+rd[itex]\varphi[/itex][itex]\hat{\varphi}[/itex]
because θ constantly equals pi/2
So you can ignore the lone sin at the end of the force vector, since it goes to one. Yet it still goes to a negative value unfortunately.
 
My apologies, we must go along the circle where y>0