- #1

xxsteelxx

- 27

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## 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!