Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Exam revision -

Verify Stokes theorem directly by explicit calculation of the surface and line integrals for the hemisphere $r=c$, with $z \geq 0$, where $F = r \times k$ and $k$ is the unit vector in the positive z-direction.

Attempt:

$$F = r \times k = r Cos(\theta) \ \hat{\phi}$$

Where $r, \theta, \phi$ are defined in the normal way.

We need to check that

$$\int^{}_{C}F.dl = \int^{}_{S}\nabla \times F.ds$$

I'm getting stuck even on the line integral.

$$dl = dr \ \hat{r} + r d\theta \ \hat{\theta} + r Sin(\theta) d\phi \ \hat{\phi}$$

$$\int^{\frac{\pi}{2}}_{0} r^2 Cos(\theta) Sin(\theta) d \phi$$

$$\frac{\pi}{2}r^2 Cos(\theta) Sin(\theta)$$

This seems wrong to me but I can't figure out why?

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

Use the fact that

$$\mathbf{r} = \sin{\theta} \cos{\phi} \mathbf{x} + \sin{\theta} \sin{\phi} \mathbf{y} + \cos{\theta} \mathbf{z}$$

$$\mathbf{\theta} = \cos{\theta} \cos{\phi} \mathbf{x} + \cos{\theta} \sin{\phi} \mathbf{y} - \sin{\theta} \mathbf{z}$$

$$\mathbf{\phi} = -\sin{\phi} \mathbf{x} + \cos{\phi} \mathbf{y}$$

to derive

$$\mathbf{F} = \mathbf{r} \times \mathbf{z} = -r \sin{\theta} \, \mathbf{\phi}$$

In cartesians

$$\mathbf{F} = y \mathbf{x} - x \mathbf{y}$$

You can show that

$$\nabla \times \mathbf{F} = -2 \mathbf{z}$$

Then the surface integral is

$$\int_S dS\: \mathbf{r} \cdot \nabla \times \mathbf{F} = -4 \pi c^2 \int_0^{\pi/2} d\theta \: \sin{\theta} \cos{\theta} = -2 \pi c^2$$

The line integral comes from the simple expression for $\mathbf{F}$ in sphericals, at $\theta = \pi/2$:

$$\oint_{\partial S} \mathbf{F}\cdot c d\phi = -c^2 \int_0^{2 \pi} d\phi = -2 \pi c^2$$

as expected from Stoke's Theorem.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.