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### Introduction to Electrodynamics

Book edition 4th edition
Author(s) David J. Griffiths
Pages 613 pages
ISBN 9780321856562

# Compute the line integral of ${\mathbit{v}}{\mathbf{=}}{\mathbf{\left(}}{\mathbit{r}}{\mathbf{}}{\mathbit{c}}{\mathbit{o}}{{\mathbit{s}}}^{{\mathbf{2}}}{\mathbit{\theta }}{\mathbf{\right)}}\stackrel{\mathbf{^}}{\mathbf{r}}{\mathbf{-}}{\mathbf{\left(}}{\mathbit{r}}{\mathbf{}}{\mathbit{c}}{\mathbit{o}}{\mathbit{s}}{\mathbit{\theta }}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}{\mathbit{\theta }}{\mathbf{\right)}}\stackrel{\mathbf{^}}{\mathbf{\theta }}{\mathbf{+}}{\mathbf{3}}{\mathbit{r}}\stackrel{\mathbf{^}}{\mathbf{\varphi }}$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]

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.

See the step by step solution

## Step 1: Describe the given information

The given vector is, $v=\left(rco{s}^{2}\theta \right)\stackrel{^}{\mathrm{r}}-\left(rcos\theta sin\theta \right)\stackrel{^}{\mathrm{\theta }}+3r\stackrel{^}{\mathrm{\varphi }}$ . 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 ${\mathbit{f}}{\mathbf{}}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{,}}{\mathbit{y}}{\mathbf{,}}{\mathbit{z}}{\mathbf{\right)}}$ over an open surface area is equal to the line integral of the function ${\mathbf{\int }}{\mathbf{\left(}}{\mathbf{\nabla }}{\mathbf{×}}{\mathbit{v}}{\mathbf{\right)}}{\mathbf{.}}{\mathbit{d}}{\mathbit{s}}{\mathbf{=}}{\mathbf{\oint }}{{\mathbf{}}}_{{\mathbf{b}}}{\mathbf{}}{\mathbit{v}}{\mathbf{.}}{\mathbit{d}}{\mathbit{l}}{\mathbf{.}}$

## 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 $v=\left(rco{s}^{2}\theta \right)\stackrel{^}{\mathrm{r}}-\left(rcos\theta sin\theta \right)\stackrel{^}{\mathrm{\theta }}+3r\stackrel{^}{\mathrm{\varphi }}$ is obtained as

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

$=\frac{1}{r\mathrm{sin}\theta }\left(3r\right)\mathrm{cos}\theta \stackrel{^}{r}+\frac{1}{r}\left(-6r\right)\stackrel{^}{\theta }+0\phantom{\rule{0ex}{0ex}}=3cot\theta \stackrel{^}{r}-\stackrel{^}{6}\theta$

The differential elemental area is $da=rdrd\theta \stackrel{^}{\varphi }$ . Substitute role="math" localid="1654664744003" $3cot\stackrel{^}{r}-\stackrel{^}{6}\theta$ for $\nabla ×v$ , into the stokes theorem $\int \left(\nabla ×v\right).d\tau =-\int v.da$

role="math" localid="1654665787372" $\int \left(\nabla ×v\right).d\tau =0+\left(\underset{0}{\overset{1}{\int }}6rdr\underset{0}{\overset{\frac{\pi }{2}}{\int }}d\varphi \right)\phantom{\rule{0ex}{0ex}}=6{{\left(\frac{{r}^{2}}{2}\right)}^{1}}_{0}\left(\varphi \right)\stackrel{\frac{\pi }{2}}{0}\phantom{\rule{0ex}{0ex}}=\left(3\right)\left(\frac{\pi }{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{3\pi }{2}...................\left(1\right)\phantom{\rule{0ex}{0ex}}$

## Step 4: Compute the right side of strokes theorem

The differential length vector is given by$dl=dr\stackrel{^}{r}+rd\theta \stackrel{^}{\theta }+r\mathrm{sin}\theta \stackrel{^}{\varphi }$ . Along the path, localid="1654672580270" $\theta =\frac{\pi }{2},\theta =0$ 0 and r varies from 0 to 1.Hence the line integral becomes,

localid="1654672141711" $\int \mathrm{v}.\mathrm{dl}=\int \left(\left(\mathrm{r}{\mathrm{cos}}^{2}\mathrm{\theta }\right)\stackrel{^}{\mathrm{r}}-\left(\mathrm{r}\mathrm{cos\theta }\mathrm{sin\theta }\right)\stackrel{^}{\mathrm{\theta }}+3\mathrm{r}\stackrel{^}{\mathrm{\varphi }}\right)\left(\mathrm{dr}\stackrel{^}{\mathrm{r}}+\mathrm{d\theta }\stackrel{^}{\mathrm{\theta }}+\mathrm{r}\mathrm{sin\theta d\varphi }\stackrel{^}{\mathrm{\varphi }}\right)\phantom{\rule{0ex}{0ex}}=\int \left(\mathrm{r}{\mathrm{cos}}^{2}\mathrm{\theta }\right)\mathrm{dr}-\left({\mathrm{r}}^{2}\mathrm{cos\theta }\mathrm{sin\theta }\right)\mathrm{d\theta }+3{\mathrm{r}}^{2}\mathrm{sin\theta d\varphi }\phantom{\rule{0ex}{0ex}}=\int \left(\mathrm{r}{\mathrm{cos}}^{2}\left(\frac{\mathrm{\pi }}{2}\right)\right)\mathrm{dr}-\left({\mathrm{r}}^{2}\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}\right)\mathrm{sin}3{\mathrm{r}}^{2}\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}\right)\right)\mathrm{d\theta }3{\mathrm{r}}^{2}\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}\right)\mathrm{d\varphi }\phantom{\rule{0ex}{0ex}}=\int 0\mathrm{dr}-0\mathrm{d\theta }+3{\mathrm{r}}^{2}\mathrm{d\varphi }$

Simplify further as

localid="1654672150719" $\int \mathrm{v}.\mathrm{dl}=\int 3{\mathrm{r}}^{2}\mathrm{d\varphi }\phantom{\rule{0ex}{0ex}}=3{\mathrm{r}}^{2}\mathrm{\varphi }\phantom{\rule{0ex}{0ex}}=3{\mathrm{r}}^{2}\left(0\right)\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}$

Along the path (ii), $\theta =\frac{\pi }{2},r=1$, and $\varphi$ varies from 0 tolocalid="1654667564073" $\frac{\pi }{2}$ .Hence the line integral becomes,

localid="1654672163671" $\int \mathrm{v}.\mathrm{dl}=\int \left(\left(\mathrm{r}{\mathrm{cos}}^{2}\mathrm{\theta }\right)\stackrel{^}{\mathrm{r}}-\left(\mathrm{r}\mathrm{cos\theta }\mathrm{sin\theta }\right)\stackrel{^}{\mathrm{\theta }}+3\mathrm{r}\stackrel{^}{\mathrm{\varphi }}\right)\left(\mathrm{dr}\stackrel{^}{\mathrm{r}}+\mathrm{d\theta }\stackrel{^}{\mathrm{\theta }}+\mathrm{r}\mathrm{sin\theta d\varphi }\stackrel{^}{\mathrm{\varphi }}\right)\phantom{\rule{0ex}{0ex}}=\int \left(\mathrm{r}{\mathrm{cos}}^{2}\mathrm{\theta }\right)\mathrm{dr}-\left({\mathrm{r}}^{2}\mathrm{cos\theta }\mathrm{sin\theta }\right)\mathrm{d\theta }+3{\mathrm{r}}^{2}\mathrm{sin\theta d\varphi }\phantom{\rule{0ex}{0ex}}=\int \left(\left(1\right){\mathrm{cos}}^{2}\left(\frac{\mathrm{\pi }}{2}\right)\right)\mathrm{dr}-\left(\left({1}^{2}\right)\mathrm{cos}\left(\frac{\mathrm{\pi }}{2}\right)\mathrm{sin}\left(\frac{\pi }{2}\right)3{\mathrm{r}}^{2}\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}\right)\right)\mathrm{d\theta }+3{\mathrm{r}}^{2}\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}\right)\mathrm{d\varphi }\phantom{\rule{0ex}{0ex}}=\int 3\mathrm{d\varphi }$

Simplify further as,

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

Along the path (iii), $\theta$varies from $\frac{\pi }{2}$ to localid="1654668630697" ${\mathrm{tan}}^{1}\left(\frac{1}{2}\right)$, localid="1654668661756" $\varphi =\frac{\pi }{2}$ and localid="1654668647962" $r=\frac{1}{\mathrm{sin}\theta }$, such that localid="1654668615795" $dr=-1\frac{1}{{\mathrm{sin}}^{2}\theta }\mathrm{cos}\theta d\theta$ Hence the line integral becomes,

localid="1654672691611" $\int \mathrm{v}.\mathrm{dl}=\int \left(\left(\mathrm{r}{\mathrm{cos}}^{2}\mathrm{\theta }\right)\stackrel{^}{\mathrm{r}}-\left(\mathrm{r}\mathrm{cos\theta }\mathrm{sin\theta }\right)\stackrel{^}{\mathrm{\theta }}+3\mathrm{r}\stackrel{^}{\mathrm{\varphi }}\right)\left(\mathrm{dr}\stackrel{^}{\mathrm{r}}+dr\stackrel{^}{r}+\mathrm{rd\theta }\stackrel{^}{\mathrm{\theta }}\mathrm{r}\mathrm{sin\theta d\varphi }\stackrel{^}{\mathrm{\varphi }}\right)\phantom{\rule{0ex}{0ex}}=\int \left(\mathrm{r}{\mathrm{cos}}^{2}\mathrm{\theta }\right)\mathrm{dr}-\left({\mathrm{r}}^{2}\mathrm{cos\theta }\mathrm{sin\theta }\right)\mathrm{d\theta }+3{\mathrm{r}}^{2}\mathrm{sin\theta d\varphi }\phantom{\rule{0ex}{0ex}}=\int \left(\frac{1}{\mathrm{sin}\theta }{\mathrm{cos}}^{2}\theta \right)\left(-\frac{1}{\mathrm{sin}\theta }\mathrm{cos}\theta d\theta \right)-\left({\left(\frac{1}{\mathrm{sin}\theta }\right)}^{2}\mathrm{cos}\theta \mathrm{sin}\theta \right)d\theta +0$

Simplify further as,

localid="1654670347504" $\int \mathrm{v}.\mathrm{dl}=-\int \left(\frac{{\mathrm{cos}}^{3}\mathrm{\theta }}{{\mathrm{sin}}^{3}\mathrm{\theta }}+\frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}\right)\mathrm{d\theta }\phantom{\rule{0ex}{0ex}}=\int \frac{\mathrm{cos\theta }}{\mathrm{sin\theta }}\left(\frac{{\mathrm{cos}}^{2}\mathrm{\theta }+{\mathrm{sin}}^{2}\mathrm{\theta }}{{\mathrm{sin}}^{2}\mathrm{\theta }}\right)\mathrm{d\theta }\phantom{\rule{0ex}{0ex}}=\underset{\frac{\mathrm{\pi }}{2}}{\overset{{\mathrm{tan}}^{-1}{\left(\frac{1}{2}\right)}^{2}}{\int }}\frac{\mathrm{cos\theta }}{{\mathrm{sin}}^{3}\mathrm{\theta }}\mathrm{d\theta }...........\left(2\right)$

Let localid="1654669903166" $x=\mathrm{sin}\theta$, then $\mathrm{dx}={\mathrm{cos}}^{2}\mathrm{\theta d\theta }$.Substitute x for localid="1654672592461" $\mathrm{sin}\theta$ and dx for $\mathrm{cos}\theta d\theta$ into equation (2)

$\int v.dl=-\int v.dl=\frac{1}{2{\mathrm{sin}}^{2}\theta }\phantom{\rule{0ex}{0ex}}=\frac{1}{2{x}^{2}}$

Substitute back for into above result as,

localid="1654672710295" $\int \mathrm{v}.\mathrm{dl}=\frac{1}{2{\mathrm{sin}}^{2}\mathrm{\theta }}$

Evaluate the limit as,

localid="1654672130422"

Along the path (iv), ${\text{θ=tan}}^{-1}{\left(\frac{1}{2}\right)}_{,}\varphi =\frac{\pi }{2}$ and r varies from localid="1654671277385" $\sqrt{5}$ to 0, Hence the line integral becomes,

localid="1654672112340" $\int \mathrm{v}.\mathrm{dl}=\int \left(\left({\mathrm{rcos}}^{2}\mathrm{\theta }\right)\right)\stackrel{^}{\mathrm{r}}-\left(\mathrm{r}\mathrm{cos\theta sin\theta }\right)\stackrel{^}{\mathrm{\theta }}+3\mathrm{r}\stackrel{^}{\mathrm{\varphi }}\right)\left(\mathrm{dr}\stackrel{^}{\mathrm{r}}+\mathrm{rd\theta }\stackrel{^}{\mathrm{\theta }}+\mathrm{rsin\theta d\varphi }\stackrel{^}{\mathrm{\varphi }}\phantom{\rule{0ex}{0ex}}=\int \left(\mathrm{r}{\mathrm{cos}}^{2}\left({\mathrm{tan}}^{-1}\left(\frac{1}{2}\right)\right)\right)\mathrm{dr}\phantom{\rule{0ex}{0ex}}=\underset{\sqrt{5}}{\overset{0}{\int }}0.8\mathrm{rdr}\phantom{\rule{0ex}{0ex}}=0.8{{\left(\frac{{\mathrm{r}}^{2}}{2}\right)}^{0}}_{\sqrt{5}}\phantom{\rule{0ex}{0ex}}$

Simplify further as,

$\int v.d=0.8\left(0-\frac{{\left(\sqrt{5}\right)}^{2}}{2}\right)\phantom{\rule{0ex}{0ex}}=-2$

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

$\oint v.dl=0+\frac{3\pi }{2}+2+\left(-2\right)\phantom{\rule{0ex}{0ex}}=\frac{3\pi }{2}.........\left(3\right)$

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

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