# Proof for my exam tomorrow

1. Dec 9, 2005

### stunner5000pt

stokes theorem says $$\int_{S} (\nabla \times G) \bullet dA = \oint_{C} G \bullet ds$$ where C is a closed curve bounding the open surface S. Note : dA =n dA, and ds = t ds

use stokes theorem to prove that hte Faraday law of induction $$\oint_{C} E \bullet t ds = -\frac{d}{dt} \int_{S_{C}} B \bullet n dA$$ can be written as a differential equation $$\nabla \times E + \frac{\partial B}{\partial t} = 0$$

now i can easily rearrange the left hand side of the integral equation to get $$\int_{S_{C}} (\nabla \times E) \bullet dA = - \frac{d}{dt} \int_{S_{C}} B \bullet n dA$$
now im not allowd to 'cancel' out the dA terms can i ? Or perhaps find the gradient on each side?? Please help on this!

Last edited: Dec 9, 2005
2. Dec 9, 2005

### StatusX

First, that first integral should be over the surface SC. Then you can bring the d/dt inside the integral on the right, and argue that since this equality must hold for any region of integration, the integrands must be equal (otherwise, you could focus the integral over the region where they weren't equal and you'd get that the integrals weren't equal).

3. Dec 9, 2005

### Cyrus

crossing out the dA's would not make sense, I hope you realize that.