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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 1.56

Here's a 3D render of the path from Figure 1.49. (Feel free to zoom out and drag around to see it from different points of view).

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

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

Problem 1.56 first asks us to do the right hand side of the equality above, and then check it using the surface integral corresponding to Stokes' theorem.

The line integral is straight forward. It consists of 3 paths on which we can deconstruct the closed path over:

$${\oint }_{\mathcal{P}}\mathbf{v}\cdot d\mathbf{l}={\int }_{L_1}\mathbf{v}\cdot d{\mathbf{r}}_1+{\int }_{L_2}\mathbf{v}\cdot d{\mathbf{r}}_2+{\int }_{L_3}\mathbf{v}\cdot d{\mathbf{r}}_3$$

Where $L_1$ is the straight path on the $y$-axis, $L_2$ the diagonal path on the $yz$plane, and $L_3$ is the straigt path on the $z$-axis. To properly express these paths, we introduce the following 3 curves

$$\begin{array}{rl}{\mathbf{r}}_1& =(0,t,0)\\ {\mathbf{r}}_2& =(0,1-t,2t)\\ {\mathbf{r}}_3& =(0,0,2-2t)\end{array}$$

All of which will go from $t_i=0$ to $t_f=1$: (notice that this is the $yz$-plane)

For which we have the differentials:

$$\begin{array}{rl}d{\mathbf{r}}_1& =(0,1,0)dt\\ d{\mathbf{r}}_2& =(0,-1,2)dt\\ d{\mathbf{r}}_3& =(0,0,-2)dt\end{array}$$

Before computing the integral, we must write the field $\mathbf{v}$ in terms of the curve parameter $t$.

For the first curve, we have

$$\begin{array}{rl}x& =0\\ y& =t\\ z& =0\end{array}$$

Thus

$$\begin{array}{rl}{\mathbf{v}}_{\text{Curve 1}}& =(6,y{z}^2,3y+z)\\ & \to (6,0,3t)\end{array}$$

For the second curve, we have

$$\begin{array}{rl}x& =0\\ y& =1-t\\ z& =2t\end{array}$$

Thus

$$\begin{array}{rl}{\mathbf{v}}_{\text{Curve 2}}& =(6,y{z}^2,3y+z)\\ & \to (6,(1-t)(2t{)}^2,3(1-t)+2t)\\ & =(6,4t^2-4t^3,3-t)\end{array}$$

For the third curve, we have

$$\begin{array}{rl}x& =0\\ y& =0\\ z& =2-2t\end{array}$$

Thus

$$\begin{array}{rl}{\mathbf{v}}_{\text{Curve 3}}& =(6,y{z}^2,3y+z)\\ & \to (6,0,2-2t)\end{array}$$

Thus, the closed line integral is:

$$\begin{array}{rl}{\oint }_{\mathcal{P}}\mathbf{v}\cdot d\mathbf{l}& ={\int }_{L_1}{\mathbf{v}}_{\text{Curve 1}}\cdot d{\mathbf{r}}_1+{\int }_{L_2}{\mathbf{v}}_{\text{Curve 2}}\cdot d{\mathbf{r}}_2+{\int }_{L_3}{\mathbf{v}}_{\text{Curve 3}}\cdot d{\mathbf{r}}_3\\ & ={\int }_0^{1}0dt+{\int }_0^1-(4t^2-4t^3)+2(3-t)dt+{\int }_0^1-2(2-2t)dt\\ & =\frac{8}{3}\end{array}$$

Let's now use Stokes' theorem and compute the surface integral of the curl of $\mathbf{v}$. At the end we should find that the surface integral is equal to $\frac{8}{3}$. It is clear that:

$$\mathrm{\nabla }\times \mathbf{v}=(3-2yz,0,0)$$

And we have

$$d\mathbf{a}=(dydz,0,0)$$

Using the diagonal path, we see that $y$ and $z$ are related by:

$$z=-2y+2$$

So the limits of integration with respect to $z$ are going to be

$$\begin{array}{rl}{z}_i& =0\\ z_f& =-2y+2\end{array}$$

And for $y$ we simply have $y_i=0$ and $y_f=1$. The surface integral is:

$$\begin{array}{rl}{\int }_{\mathcal{S}}(\mathrm{\nabla }\times \mathbf{v})\cdot d\mathbf{a}& ={\int }_{y=0}^1{\int }_{z=0}^{-2y+2}3-2yzdzdy\\ & ={\int }_{y=0}^{1}3(2-2y)-(2-2y{)}^{2}ydy\\ & =\frac{8}{3}\end{array}$$

As expected.

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