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]

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

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 $\mathbf{\int }\mathbf{(}\mathbf{\nabla }\mathbf{\times }\mathit{v}\mathbf{)}\mathbf{.}\mathit{d}\mathit{s}\mathbf{=}\mathbf{\oint }{\mathbf{}}_{\mathbf{b}}\mathbf{}\mathit{v}\mathbf{.}\mathit{d}\mathit{l}\mathbf{.}$

The formula of curl of a vector in spherical coordinates is

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

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

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

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

role="math" localid="1654665787372" $$\begin{gathered} \int (\nabla \times v).d\tau =0+\left(\underset{0}{\overset{1}{\int }}6rdr\underset{0}{\overset{\frac{\pi }{2}}{\int }}d\varphi \right) \\ =6{{\left(\frac{r^2}{2}\right)}^1}_0\left(\varphi \right)\stackrel{\frac{\pi }{2}}{0} \\ =\left(3\right)\left(\frac{\pi }{2}\right) \\ =\frac{3\pi }{2}...................\left(1\right) \end{gathered}$$

The differential length vector is given by$dl=dr\hat{r}+rd\theta \hat{\theta }+r\sin \theta \hat{\varphi }$ . 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" $$\begin{gathered} \int \mathrm{v}.\mathrm{dl}=\int \left((\mathrm{r}{\cos }^2\mathrm{\theta }\right)\hat{\mathrm{r}}-(\mathrm{r}\mathrm{cos\theta }\mathrm{sin\theta })\hat{\mathrm{\theta }}+3\mathrm{r}\hat{\mathrm{\varphi }}\left)\right(\mathrm{dr}\hat{\mathrm{r}}+\mathrm{d\theta }\hat{\mathrm{\theta }}+\mathrm{r}\mathrm{sin\theta d\varphi }\hat{\mathrm{\varphi }}) \\ =\int (\mathrm{r}{\cos }^2\mathrm{\theta })\mathrm{dr}-({\mathrm{r}}^2\mathrm{cos\theta }\mathrm{sin\theta })\mathrm{d\theta }+3{\mathrm{r}}^2\mathrm{sin\theta d\varphi } \\ =\int \left(\mathrm{r}{\cos }^2\left(\frac{\mathrm{\pi }}{2}\right)\right)\mathrm{dr}-\left({\mathrm{r}}^2\cos \left(\frac{\mathrm{\pi }}{2}\right)\sin 3{\mathrm{r}}^2\sin \left(\frac{\mathrm{\pi }}{2}\right)\right)\mathrm{d\theta }3{\mathrm{r}}^2\sin \left(\frac{\mathrm{\pi }}{2}\right)\mathrm{d\varphi } \\ =\int 0\mathrm{dr}-0\mathrm{d\theta }+3{\mathrm{r}}^2\mathrm{d\varphi } \end{gathered}$$

Simplify further as

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

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

localid="1654672163671" $$\begin{gathered} \int \mathrm{v}.\mathrm{dl}=\int \left((\mathrm{r}{\cos }^2\mathrm{\theta }\right)\hat{\mathrm{r}}-(\mathrm{r}\mathrm{cos\theta }\mathrm{sin\theta })\hat{\mathrm{\theta }}+3\mathrm{r}\hat{\mathrm{\varphi }}\left)\right(\mathrm{dr}\hat{\mathrm{r}}+\mathrm{d\theta }\hat{\mathrm{\theta }}+\mathrm{r}\mathrm{sin\theta d\varphi }\hat{\mathrm{\varphi }}) \\ =\int (\mathrm{r}{\cos }^2\mathrm{\theta })\mathrm{dr}-({\mathrm{r}}^2\mathrm{cos\theta }\mathrm{sin\theta })\mathrm{d\theta }+3{\mathrm{r}}^2\mathrm{sin\theta d\varphi } \\ =\int (\left(1\right){\cos }^2\left(\frac{\mathrm{\pi }}{2}\right))\mathrm{dr}-(\left(1^2\right)\cos \left(\frac{\mathrm{\pi }}{2}\right)\sin \left(\frac{\pi }{2}\right)3{\mathrm{r}}^2\sin \left(\frac{\mathrm{\pi }}{2}\right))\mathrm{d\theta }+3{\mathrm{r}}^2\sin (\frac{\mathrm{\pi }}{2})\mathrm{d\varphi } \\ =\int 3\mathrm{d\varphi } \end{gathered}$$

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

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

Simplify further as,

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

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

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

Evaluate the limit as,

localid="1654672130422" $$\begin{gathered} \int v.dl={{\left(\frac{1}{2{\sin }^2\theta }\right)}_{\frac{x}{2}}}^{{\tan }^{-1}\left(\frac{1}{2}\right)} \\ =\frac{1}{2{\sin }^2\left({\tan }^{-1}\left({\displaystyle \frac{1}{2}}\right)\right)}-\frac{1}{2{\sin }^2\left({\displaystyle \frac{\pi }{2}}\right)} \\ =\frac{1}{2(0.2)}-\frac{1}{2} \\ =2 \end{gathered}$$

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

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

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.