How do I evaluate this integral of a closed surface?

Question

Suppose I have some vector field $\mathbf{E} = (-ye^{-2t}, xe^{-2t}, 0)$, and vector field $\mathbf{B} = (0,0,e^{-2t})$ that depends on time t. I want to evaluate

$$\int_S \mathbf{B}\cdot d\mathbf{S}$$ and

$$\int_C \mathbf{E}\cdot d\mathbf{x}$$

where $S = x^2 + y^2 < 1$ and $z = 0$, and C is the curve bounding S. Then I want to finally show that

$$\int_C \mathbf{E}\cdot d\mathbf{x} = -\frac{d}{dt} \int_S \mathbf{B}\cdot d\mathbf{S}$$

I am at the first step of showing $\int_S \mathbf{B}\cdot d\mathbf{S}$ to be equal to zero by Gauss's theorem, since $\int_S \mathbf{B}\cdot d\mathbf{S} = \int_v (\nabla \cdot \mathbf{F}) dV = 0$, and for the disc, there is no volume element available. Is this the correct application of the theorem?

For the next step, I am showing that $\int_C \mathbf{E}\cdot d\mathbf{x}$ is effectively $\int_C e^{-2t} dz$.

However, how do I combine all the above information to prove the last step?

Answer 1

First, I recommend you to describe your geometry with cylindrical geometr, problem become much easier. Let's start computing your differential surface $d\mathbf{S}$, and differential contour $d\mathbb{x}$:

$$d\mathbf{S}=(0,0,1)^TdS=(0,0,1)^T \rho d\theta d\rho$$

$$d\mathbf{x}=(-\sin(\theta),\cos(\theta),0)^T\rho d\theta$$

So, $\int_S \mathbf{B}\cdot d\mathbf{S}$ can be rewritten as:

$$\int_S \mathbf{B}\cdot d\mathbf{S} = \int_{0}^{2\pi} \int_{0}^{1} e^{-2t}\rho d\rho d\theta = \pi e^{-2t}$$

While the contour integral becomes (taking into account that $x=\rho \cos(\theta)$ and $y=\rho \sin(\theta)$):

$$\int_C \mathbf{E}\cdot d\mathbf{x} = \int_0^{2\pi} (\sin^2(\theta)+\cos^2(\theta))e^{2t}1d\theta=2\pi e^{-2t}$$

Now, one can easily check that:

$$\frac{d}{dt}\int_S \mathbf{B}\cdot d\mathbf{S} = -2\pi e^{-2t} = -\int_C \mathbf{E}\cdot d\mathbf{x}$$

Answer 2

• $S$ is not closed—it is not the boundary of a compact volume. Therefore, you may not use Gauss's theorem to evaluate the magnetic flux $\int_S\mathbf{B}\cdot\mathrm{d}\mathbf{S}$.
• Neither $\mathbf{E}$ nor the line element of $C$ have any component in the $\mathbf{\hat{z}}$ direction, so your expression for the electromotivity $\int_C\mathbf{E}\cdot\mathrm{d}\mathbf{x}$ cannot reduce to $\int_C\mathrm{e}^{-2t}\mathrm{d}z$.

Instead, you should evaluate these expressions using cylindrical coordinates $(s,\phi,z)$: $$\begin{align}\mathbf{E}&=\mathrm{e}^{-2t}\boldsymbol{\hat{\phi}}&\mathbf{B}=\mathrm{e}^{-2t}\mathbf{\hat{z}}\\ \int_{C}\mathbf{E}\cdot\mathrm{d}\mathbf{x}&=\int_0^{2\pi}\mathrm{e}^{-2t}\mathrm{d}\phi&\int_{S}\mathbf{B}\cdot\mathrm{d}\mathbf{S}&=\int_0^{2\pi} \int_0^1\mathrm{e}^{-2t}s\mathrm{d}s\,\mathrm{d}\phi\text{;}\end{align}$$ then you can verify Faraday's induction law $\int_{C}\mathbf{E}\cdot\mathrm{d}\mathbf{x}=-\frac{\mathrm{d}}{\mathrm{d}t}\int_{S}\mathbf{B}\cdot\mathrm{d}\mathbf{S}$ directly.