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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 1.55

Let's recall The Fundamental Theorem for curls, also known as Stokes' theorem:

$$\begin{array}{}\text{(Eq. 1.57, page 33)}& {\int }_{\mathcal{S}}(\mathrm{\nabla }\times \mathbf{v})\cdot d\mathbf{a}={\oint }_{\mathcal{P}}\mathbf{v}\cdot d\mathbf{l}\end{array}$$

Given the circular path mentioned above, the left hand side's surface integral will be over the surface of the circle of radius $R$, while the line integral will be over its circumference.

The field to be integrated over the entire area is the curl of $\mathbf{v}$:

$$\mathrm{\nabla }\times \mathbf{v}=(-a+b)\hat{\mathbf{z}}$$

This is a constant vector pointing perpendicularly in or out (depending on the vaulues of $a$ and $b$) our of the circle's surface.

The line integral will be withr respect to the $\mathbf{v}$ field itself. Here's a live demonstration of the $\mathbf{v}$ field. Feel free to use the sliders to modify the field.

Let's now calculate both integrals and compare the results.

Starting with the surface integral:

$$\begin{array}{rl}{\int }_{\mathcal{S}}(\mathrm{\nabla }\times \mathbf{v})\cdot d\mathbf{a}& ={\int }_{\mathcal{S}}(-a+b)\hat{\mathbf{z}}\cdot da\hat{\mathbf{z}}\\ & ={\int }_{\mathcal{S}}(-a+b)da\\ & =(-a+b){\int }_{\mathcal{S}}da\\ & =(-a+b)\cdot \left(\text{area of a circle of radius }R\right)\\ & =(-a+b)\pi R^2\end{array}$$

It is true that $d\mathbf{a}=dxdy\hat{\mathbf{z}}$, but keeping it ``symbolic'' can be helpful (in this particular problem) to show that the integral is just the are of the surface. Longer ways to solve this integral include actually writing $dxdy$ and writing the limits of integration to bound the area to a circle of radius $R$, or converting the area element into a polar area element (which then simplifies the integration limits).

Let's now do the line integral. A better way to think about line integrals is to think about the following expression:

$${\int }_{\mathcal{C}}\mathbf{v}\cdot d\mathbf{r}$$

Where $\mathcal{C}$ is a Curve, a fancy word for a path. This curve depends on a single parameter which we usually denote with $t$. You can think of a curve as a vector that depends on a single variable.

Notice that a circle of radius $R$ can be represented by the following curve:

$$\text{Circle of radius }R=\{\left(\begin{array}{c}R\cos \left(t\right)\\ R\sin \left(t\right)\end{array}\right)|t\in [0,2\pi ]\}$$

That is, we will start to only deal with the variable $t$ rather than $x$ and $y$ itself. Using this variable, we see that the vector field $\mathbf{v}$ can be rewritten as:

$$\mathbf{v}=\left(\begin{array}{c}ay\\ bx\end{array}\right)\to \left(\begin{array}{c}aR\sin \left(t\right)\\ bR\cos \left(t\right)\end{array}\right)$$

Notice then that for a given $t$, we now know how to find the Cartesian point of the curve AND the components of the vector field at that point on the curve:

And we have the infinitesimal:

$$\begin{array}{rl}d\mathbf{r}& =\frac{d\mathbf{r}}{dt}dt\\ & =\left(\begin{array}{c}-R\sin \left(t\right)\\ R\cos \left(t\right)\end{array}\right)dt\end{array}$$

So the line integral is:

$$\begin{array}{rl}\int \mathbf{v}\cdot d\mathbf{r}& ={\int }_0^{2\pi }\left(\begin{array}{c}aR\sin \left(t\right)\\ bR\cos \left(t\right)\end{array}\right)\cdot \left(\begin{array}{c}-R\sin \left(t\right)\\ R\cos \left(t\right)\end{array}\right)dt\\ & ={\int }_0^{2\pi }-a{R}^2{\sin }^2(t)+b{R}^2{\cos }^2(t)dt\\ & ={\int }_0^{2\pi }-a{R}^2{\sin }^2(t)dt+{\int }_0^{2\pi }b{R}^2{\cos }^2(t)dt\\ & =-a{R}^2{\int }_0^{2\pi }{\sin }^{2}tdt+b{R}^2{\int }_0^{2\pi }{\cos }^2(t)dt\\ & =-a{R}^2\left(\pi \right)+b{R}^2\left(\pi \right)\\ & =(-a+b)\pi R^2\end{array}$$

Which is precisely the result we obtained by performing the surface integral of the curl of the field.

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