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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 1.57

Here's a 3D render of Figure 1.50:

Let's first calculate the line integral

The line integral can be split into 4 separate parts:

From $D$ to $A$, from $A$ to $B$, from $B$ to $C$, and from $C$ to $D$. (As seen in the 3D render above).

The first 2 (from $D$ to $A$ and from $A$ to $B$) can be computed trivially in spherical coordiantes.

The last 2 (from $B$ to $C$ and from $C$ to $D$), although possible in spherical, are trivial if we consider the vector field $\mathbf{v}$ in cylindrical coordinates.

First line integral:

We have a path in the $\hat{r}$-direction, meaning that move at constant $\theta$ and $\varphi$, that is:

$$d\theta =d\varphi =0$$

Further, along this path this constant variables have the values

$$\theta =\frac{\pi }{2}|\varphi =0$$

The line integral is:

$$\begin{array}{rl}\int \mathbf{v}\cdot d\mathbf{l}& =\int \mathbf{v}\cdot (dr,rd\theta ,r\sin \left(\theta \right)d\varphi )\\ & ={\int }_{r_1}^{r_2}{v}_{r}dr\\ & ={\int }_0^{1}r{\cos }^2(\theta )dr\\ & ={\cos }^2(\theta ){\int }_0^{1}rdr\\ & =\cancel{{\cos }^2(\pi /2)}{\int }_0^{1}rdr\\ \text{(Since }\cos \left(\pi /2\right)=0\text{)}& & =0\end{array}$$

Second line integral:

We have a path in the $\hat{\varphi }$-direction, meaning that move at constant $r$ and $\theta$, that is:

$$dr=d\theta =0$$

Further, along this path this constant variables have the values

$$r=1|\theta =\frac{\pi }{2}$$

The line integral is:

$$\begin{array}{rl}\int \mathbf{v}\cdot d\mathbf{l}& =\int \mathbf{v}\cdot (dr,rd\theta ,r\sin \left(\theta \right)d\varphi )\\ & ={\int }_{{\varphi }_1}^{{\varphi }_2}{v}_{\varphi }(r\sin \left(\theta \right))d\varphi \\ & ={\int }_0^{\pi /2}3r^2\sin \theta d\varphi \\ & =3(1{)}^2\sin (\pi /2){\int }_0^{\pi /2}d\varphi \\ & =3(\frac{\pi }{2}-0)\\ & =\frac{3}{2}\pi \end{array}$$

The $\mathbf{v}$ field in cylindrical coordinates:

Read Prologue to 1.57: A brief discussion on conversions between spherical and cylidrical coordinates if you want to learn more details about this derivation.

$$\begin{array}{rl}\mathbf{v}& =(r{\cos }^2\theta )\hat{\mathbf{r}}-(r\cos \left(\theta \right)\sin \left(\theta \right))\hat{\theta }+3r\hat{\varphi }\\ & =\sqrt{s^2+z^2}\frac{z^2}{s^2+z^2}\hat{\mathbf{r}}-\sqrt{s^2+z^2}\frac{z}{\sqrt{s^2+z^2}}\sqrt{1-\frac{z^2}{s^2+z^2}}\hat{\theta }+3\sqrt{s^2+z^2}\hat{\varphi }\\ & =\frac{z^2}{\sqrt{s^2+z^2}}\hat{\mathbf{r}}-z\sqrt{1-\frac{z^2}{s^2+z^2}}\hat{\theta }+3\sqrt{s^2+z^2}\hat{\varphi }\\ & =z\hat{z}+3\sqrt{s^2+z^2}\hat{\varphi }\end{array}$$

This is extremely enlightening! This tells us that the $\mathbf{v}$ will never point radially out of the $z$-field. (This is an example of how changing coordinate systems might reveal geometric details about our vector field).

Third line integral:

We are not moving on the $\hat{\varphi }$-direction, only on the $z$-direction, so the integral is quite simple!

$${\int }_0^{2}zdz=2$$

Fourth line integral:

Once again, the integral is simple since we are not moving on the $\hat{\varphi }$-direction, but only on the $z$-direction.

$${\int }_2^{0}zdz=-2$$

Line integral conclusions:

$$\begin{array}{rl}\int \mathbf{v}\cdot d\mathbf{l}& =0+\frac{3\pi }{2}+2-2\\ & =\frac{3\pi }{2}\end{array}$$

Now that we computed all 4 segments of the path, we see that

Confirm result using Stokes' theorem:

Recall Stokes' theorem:

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

The curl of the field is (in spherical)

$$\mathrm{\nabla }\times \mathbf{v}=3\cot \theta \hat{r}-6\hat{\theta }+0\hat{\varphi }$$

We can immediately tell that the surface integral over the triangular surface (on the $yz$-plane) is 0, since we don't have curl in the $\varphi$-direction. For the surface integral on the $xy$-plane (the 1/4th of a circle) is given by:

$$\begin{array}{rl}\int \left(\begin{array}{ccc}3\cot \theta & -6& 0\end{array}\right)\cdot d\mathbf{a}& ={\int }_{r=0}^1{\int }_{\theta =0}^{\pi /2}6r\sin \left(\theta \right)d\theta dr\\ & =6(\frac{1}{2}-0)(\frac{\pi }{2}-0)\\ & =\frac{3\pi }{2}\end{array}$$

Agreeing with the line integral result.

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