# Work done along a semicircle

1. Dec 14, 2011

### HeisenbergW

1. The problem statement, all variables and given/known data
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$\hat{r}$ + r3*cosφ*cos(2φ) $\hat{φ}$

2. Relevant equations
Work
W=∫F*dr
where dr=dr= dr$\hat{r}$ + rsinφdφ$\hat{φ}$

3. 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$\hat{r}$ + 8*cos4φ*cos(2φ) $\hat{φ}$

W=∫F*dr

So now i dot product this with the dr, which is now dr= dr$\hat{r}$ + 2cosφ
sinφdφ$\hat{φ}$ 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 in to 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

2. Dec 15, 2011

### ehild

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

ehild

3. Dec 15, 2011

### HeisenbergW

For d$\vec{r}$ 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$\hat{r}$+rsinθd$\varphi$$\hat{\varphi}$.
Which of course just simplifies to
dr$\hat{r}$+rd$\varphi$$\hat{\varphi}$
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.

4. Dec 15, 2011

### ehild

The integration path is not clear: You can reach from (0.0,0) to (2,0,0) along the upper semicircle or the bottom one, yielding integrals of opposite signs.

ehild

5. Dec 15, 2011

### HeisenbergW

My apologies, we must go along the circle where y>0

6. Dec 15, 2011

### ehild

The integral is negative then.

ehild