Can F be expressed as the gradient of a scalar?

[xxsteelxx]

Homework Statement

Assume a vector field:[tex]\textbf{F} = \widehat{r} 2r sin\phi + \widehat{\phi} r^2 cos\phi[/tex]

a) Verify the Stokes's theorem over the ABCD contour shown in Fig. 1 .
b) Can F be expressed as the gradient of a scalar? Explain

My problems results in not being able to verify Stoke's Theorem

Homework Equations

Stoke's Theorem
[tex]\oint_{C} \textbf{F}\cdot \overrightarrow{dl} = \int \int_{S} (\nabla\times\textbf{F})\cdot \overrightarrow{dS}[/tex]

The Attempt at a Solution

We can see that the line integral and surface integrals can be dealt with in cylindrical coordinates.
for line integrals:
[tex]\int _{DA}+\int _{AB}+\int _{BC}+\int _{CD} =-\int_{r=2}^{1}2rsin(0)dr + \int_{\phi=0}^{\pi/3}cos(\phi)d\phi + \int_{r=1}^{2}2rsin(\pi/3)dr + \int_{\phi=\pi/3}^{0}4cos(\phi)d\phi[/tex]
[tex]= \frac{\sqrt3}{2} +\frac{\sqrt3}{2}(4-1) -4\frac{\sqrt3}{2}=0 [/tex]

I also obtain that [tex]\nabla\cdot \textbf{F}=\widehat{z}cos\phi(3r-2)[/tex]
And surface integral is evaluated as [tex]-\int_{r=1}^{2}\int _{\phi=0}^{\phi/3}cos\phi(3r-2)rdrd\phi=-(8-4)\frac{\sqrt3}{2}=-4\frac{\sqrt3}{2}[/tex]
The negative is due to the fact that ds is in the -z direction.
Am I doing something wrong while integrating?

Thanks in advance!

 

[SteamKing]

In your line integrals for segments AB and CD, d[itex]\phi[/itex] represents a change in angle, which s not the same as a change in arc length, dl.

 

[xxsteelxx]

True, but isn't [tex]\overrightarrow{dl} = \widehat{\phi}r d\phi [/tex] for path AB and [tex]\overrightarrow{dl} = -\widehat{\phi} r d\phi[/tex] for path CD? The after dot product, I obtain the integrals previously?

 

[HallsofIvy]

SteamKings point is that, in the integral from C to D, [itex]ds= 2d\phi[/itex]. You need another factor of 2.

 

[xxsteelxx]

Ahh I see now I was missing r = 2 in the the CD line integral, now Stoke's theorem can be verified. Thank you very much guys!