Q57P - Introduction to Electrodynamics

Question

Compute the line integral of

  $\mathit{v}\mathbf{=}\mathbf{(}\mathit{r}\mathbf{ }\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{)}\hat{\mathbf{r}}\mathbf{-}\mathbf{(}\mathit{r}\mathbf{ }\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{)}\hat{\mathbf{\theta }}\mathbf{+}\mathbf{3}\mathit{r}\hat{\mathbf{\varphi }}$

around the path shown in Fig. 1.50 (the points are labeled by their Cartesian coordinates).Do it either in cylindrical or in spherical coordinates. Check your answer, using Stokes' theorem. [Answer: 3rr /2]

                                   

Step-by-Step Solution

Answer

The line integral is evaluated to be $\frac{3\pi }{2}$ . The left and right side of the Stokes theorem gives same result. Hence strokes theorem is verified.

Step 1: Describe the given information

The given vector is, $v=(r co{s}^2\theta )\hat{\mathrm{r}}-(r cos\theta sin\theta )\hat{\mathrm{\theta }}+3r\hat{\mathrm{\varphi }}$ . The line integral of the given vector has to be evaluated over the path drawn as follows:

Step 2: Define the Stokes theorem

The integral of curl of a function $\mathit{f}\mathbf{ }\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{,}\mathit{z}\mathbf{)}$ over an open surface area is equal to the line integral of the function $\mathbf{\int }\mathbf{(}\mathbf{\nabla }\mathbf{\times }\mathit{v}\mathbf{)}\mathbf{.}\mathit{d}\mathit{s}\mathbf{=}\mathbf{\oint }{\mathbf{ }}_{\mathbf{b}}\mathbf{ }\mathit{v}\mathbf{.}\mathit{d}\mathit{l}\mathbf{.}$

Step 3: Compute the left side of strokes theorem

The formula of curl of a vector in spherical coordinates is

The curl of the vector $v=(r co{s}^2\theta )\hat{\mathrm{r}}-(r cos\theta sin\theta )\hat{\mathrm{\theta }}+3r\hat{\mathrm{\varphi }}$  is obtained as 

$\nabla \times v=\left[\begin{array}{c}\frac{1}{r \sin \theta }\left(\frac{\partial \left(\sin \theta \right(3r\left)\right)}{\partial \theta }-\frac{\partial r\cos \theta \sin \theta }{\partial \varphi }\right)\hat{r}\\ +\frac{1}{r}\left(\frac{1}{\sin \theta }\frac{\partial r {\cos }^2\theta }{\partial \varphi }-\frac{\partial \left(r\right(3r\left)\right)}{\partial r}\right)\hat{\theta }+\frac{1}{r}\left(\frac{\partial \left(r\right(r \cos \theta \sin \theta }{\partial r}-\frac{\partial (r {\cos }^2\theta )}{\partial \theta }\right)\hat{\varphi }\end{array}\right]\begin{array}{}\end{array}$

         $$\begin{gathered} =\frac{1}{r\sin \theta }\left(3r\right)\cos \theta \hat{r}+\frac{1}{r}(-6r)\hat{\theta }+0 \\ =3 cot\theta \hat{r}-\hat{6}\theta \end{gathered}$$

The differential elemental area is $da=rdrd\theta \hat{\varphi }$  . Substitute   $3cot \hat{r}- \hat{6}\theta$ for $\nabla \times v$ , into the stokes theorem $\int (\nabla \times v).d\tau =-\int v.da$

$$\begin{gathered} \int (\nabla \times v).d\tau =0+\left(\underset{0}{\overset{1}{\int }} 6rdr\underset{ 0}{\overset{\frac{\pi }{2}}{\int }} d\varphi \right) \\ =6{{\left(\frac{r^2}{2}\right)}^1}_0\left(\varphi \right) \stackrel{\frac{\pi }{2}}{0} \\ =\left(3\right)\left(\frac{\pi }{2}\right) \\ =\frac{3\pi }{2}...................\left(1\right) \end{gathered}$$

Step 4: Compute the right side of strokes theorem

The differential length vector is given by$dl=dr \hat{r}+rd\theta \hat{\theta }+r \sin \theta \hat{\varphi }$  . Along the path, $\theta =\frac{\pi }{2},\theta =0$  0 and r varies from 0 to 1.Hence the line integral becomes,

$$\begin{gathered} \int \mathrm{v}.\mathrm{dl}=\int \left((\mathrm{r} {\cos }^2 \mathrm{\theta }\right)\hat{\mathrm{r}}-(\mathrm{r} \mathrm{cos\theta } \mathrm{sin\theta })\hat{\mathrm{\theta }}+3\mathrm{r} \hat{\mathrm{\varphi }}\left)\right(\mathrm{dr} \hat{\mathrm{r}}+\mathrm{d\theta } \hat{\mathrm{\theta }}+\mathrm{r} \mathrm{sin\theta d\varphi }\hat{\mathrm{\varphi }}) \\ =\int (\mathrm{r} {\cos }^2\mathrm{\theta })\mathrm{dr}-({\mathrm{r}}^2 \mathrm{cos\theta } \mathrm{sin\theta })\mathrm{d\theta }+3{\mathrm{r}}^2\mathrm{sin\theta d\varphi } \\ =\int \left(\mathrm{r} {\cos }^2\left(\frac{\mathrm{\pi }}{2}\right)\right)\mathrm{dr}-\left({\mathrm{r}}^2 \cos \left(\frac{\mathrm{\pi }}{2}\right)\sin 3{\mathrm{r}}^2 \sin \left(\frac{\mathrm{\pi }}{2}\right)\right)\mathrm{d\theta }3{\mathrm{r}}^2\sin \left(\frac{\mathrm{\pi }}{2}\right)\mathrm{d\varphi } \\ =\int 0 \mathrm{dr}-0\mathrm{d\theta }+3{\mathrm{r}}^2\mathrm{d\varphi } \end{gathered}$$

Simplify further as

 $$\begin{gathered} \int \mathrm{v}.\mathrm{dl}=\int 3{\mathrm{r}}^2 \mathrm{d\varphi } \\ =3{\mathrm{r}}^2\mathrm{\varphi } \\ =3{\mathrm{r}}^2\left(0\right) \\ =0 \end{gathered}$$

Along the path (ii), $\theta =\frac{\pi }{2} ,r=1$, and $\varphi$ varies from 0 to$\frac{\pi }{2}$ .Hence the line integral becomes,

$$\begin{gathered} \int \mathrm{v}.\mathrm{dl}=\int \left((\mathrm{r} {\cos }^2 \mathrm{\theta }\right)\hat{\mathrm{r}}-(\mathrm{r} \mathrm{cos\theta } \mathrm{sin\theta })\hat{\mathrm{\theta }}+3\mathrm{r} \hat{\mathrm{\varphi }}\left)\right(\mathrm{dr} \hat{\mathrm{r}}+\mathrm{d\theta } \hat{\mathrm{\theta }}+\mathrm{r} \mathrm{sin\theta d\varphi }\hat{\mathrm{\varphi }}) \\ =\int (\mathrm{r} {\cos }^2\mathrm{\theta })\mathrm{dr}-({\mathrm{r}}^2 \mathrm{cos\theta } \mathrm{sin\theta })\mathrm{d\theta }+3{\mathrm{r}}^2\mathrm{sin\theta d\varphi } \\ =\int (\left(1\right){\cos }^2\left(\frac{\mathrm{\pi }}{2}\right))\mathrm{dr}-(\left(1^2\right) \cos \left(\frac{\mathrm{\pi }}{2}\right)\sin \left(\frac{\pi }{2}\right)3{\mathrm{r}}^2 \sin \left(\frac{\mathrm{\pi }}{2}\right))\mathrm{d\theta }+3{\mathrm{r}}^2\sin (\frac{\mathrm{\pi }}{2})\mathrm{d\varphi } \\ =\int 3\mathrm{d\varphi } \end{gathered}$$

Simplify further as,

$$\begin{gathered} \int \mathrm{v}.\mathrm{dl}=\underset{0}{\overset{\frac{\pi }{2}}{\int }}3\mathrm{d\varphi } \\ =3\left(\varphi \right)\stackrel{\frac{\pi }{2}}{0} \\ =\frac{3\pi }{2} \end{gathered}$$

Along the path (iii),  $\theta$varies from  $\frac{\pi }{2}$ to  ${\tan }^1\left(\frac{1}{2}\right)$, $\varphi =\frac{\pi }{2}$  and  $r=\frac{1}{\sin \theta }$, such that $dr=-1\frac{1}{{\sin }^2\theta }\cos \theta d\theta$ Hence the line integral becomes,

$$\begin{gathered} \int \mathrm{v}.\mathrm{dl}=\int \left((\mathrm{r} {\cos }^2 \mathrm{\theta }\right)\hat{\mathrm{r}}-(\mathrm{r} \mathrm{cos\theta } \mathrm{sin\theta })\hat{\mathrm{\theta }}+3\mathrm{r} \hat{\mathrm{\varphi }}\left)\right(\mathrm{dr} \hat{\mathrm{r}}+dr \hat{r}+\mathrm{rd\theta } \hat{\mathrm{\theta }} \mathrm{r} \mathrm{sin\theta d\varphi }\hat{\mathrm{\varphi }}) \\ =\int (\mathrm{r} {\cos }^2\mathrm{\theta })\mathrm{dr}-({\mathrm{r}}^2 \mathrm{cos\theta } \mathrm{sin\theta })\mathrm{d\theta }+3{\mathrm{r}}^2\mathrm{sin\theta d\varphi } \\ =\int (\frac{1}{\sin \theta }{\cos }^2\theta \left)\right(-\frac{1}{\sin \theta }\cos \theta d\theta )-({\left(\frac{1}{\sin \theta }\right)}^2\cos \theta \sin \theta )d\theta +0 \end{gathered}$$

Simplify further as,

 $$\begin{gathered} \int \mathrm{v}.\mathrm{dl}=-\int \left(\frac{{\cos }^3\mathrm{\theta }}{{\sin }^3\mathrm{\theta }}+\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}\right)\mathrm{d\theta } \\ =\int \frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}\left(\frac{{\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }}{{\sin }^2\mathrm{\theta }}\right)\mathrm{d\theta } \\ =\underset{\frac{\mathrm{\pi }}{2}}{\overset{{\tan }^{-1}{\left(\frac{1}{2}\right)}^2}{\int }}\frac{\mathrm{cos\theta }}{{\sin }^3\mathrm{\theta }}\mathrm{d\theta } ...........\left(2\right) \end{gathered}$$

Let  $x=\sin \theta$, then $\mathrm{dx}={\cos }^2\mathrm{\theta d\theta }$.Substitute x  for $\sin \theta$ and dx  for  $\cos \theta d\theta$ into equation (2)

 $$\begin{gathered} \int v.dl=-\int v.dl=\frac{1}{2{\sin }^2\theta } \\ =\frac{1}{2x^2} \end{gathered}$$

Substitute back   for  into above result as,

 $\int \mathrm{v}.\mathrm{dl}=\frac{1}{2{\sin }^2 \mathrm{\theta }}$

 Evaluate the limit as,

$$\begin{gathered} \int v.dl={{\left(\frac{1}{2{\sin }^2\theta }\right)}_{\frac{x}{2}}}^{{\tan }^{-1}\left(\frac{1}{2}\right)} \\ =\frac{1}{2{\sin }^2\left({\tan }^{-1}\left({\displaystyle \frac{1}{2}}\right)\right)}-\frac{1}{2{\sin }^2\left({\displaystyle \frac{\pi }{2}}\right)} \\ =\frac{1}{2(0.2)}-\frac{1}{2} \\ =2 \end{gathered}$$

 

Along the path (iv), ${\text{θ=tan}}^{-1}{\left(\frac{1}{2}\right)}_{,}\varphi =\frac{\pi }{2}$ and r   varies from $\sqrt{5}$  to 0, Hence the line integral becomes,

$$\begin{gathered} \int \mathrm{v}.\mathrm{dl}=\int \left(\left({\mathrm{rcos}}^2\mathrm{\theta }\right)\right)\hat{\mathrm{r}}-(\mathrm{r} \mathrm{cos\theta sin\theta })\hat{\mathrm{\theta }}+3\mathrm{r}\hat{\mathrm{\varphi }}\left)\right(\mathrm{dr} \hat{\mathrm{r}}+\mathrm{rd\theta } \hat{\mathrm{\theta }} +\mathrm{rsin\theta d\varphi }\hat{\mathrm{\varphi }} \\ =\int \left(\mathrm{r} {\cos }^2\left({\tan }^{-1}\left(\frac{1}{2}\right)\right)\right)\mathrm{dr} \\ =\underset{\sqrt{5}}{\overset{0}{\int }}0.8\mathrm{rdr} \\ =0.8{{\left(\frac{{\mathrm{r}}^2}{2}\right)}^0}_{\sqrt{5}} \end{gathered}$$

Simplify further as,

 $$\begin{gathered} \int v.d=0.8\left(0-\frac{{\left(\sqrt{5}\right)}^2}{2}\right) \\ =-2 \end{gathered}$$

The integral of all the four parts are added to give:

     $$\begin{gathered} \oint v.dl=0+\frac{3\pi }{2}+2+\left(-2\right) \\ =\frac{3\pi }{2} .........\left(3\right) \end{gathered}$$                           

 

From equation (1) and (3), the left and right side gives same result. Hence strokes theorem is verified.