91Ó°ÊÓ

The given vector is, $v=(rco{s}^2\mathrm{θ})\hat{\mathrm{r}}-(rcos\mathrm{θ}sin\mathrm{θ})\widehat{\mathrm{θ}}+3r\widehat{\mathrm{ϕ}}$. 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 $\mathbf{∫}\mathbf{(}\mathbf{∇}\mathbf{×}\mathit{v}\mathbf{)}\mathbf{.}\mathit{d}\mathit{s}\mathbf{=}\mathbf{âˆ\circledR }{\mathbf{}}_{\mathbf{b}}\mathbf{}\mathit{v}\mathbf{.}\mathit{d}\mathit{l}\mathbf{.}$

The formula of curl of a vector in spherical coordinates is

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

$∇×v=\left[\begin{array}{c}\frac{1}{r\sin \mathrm{θ}}\left(\frac{∂\left(\sin \mathrm{θ}\right(3r\left)\right)}{∂\mathrm{θ}}-\frac{∂r\cos \mathrm{θ}\sin \mathrm{θ}}{∂\mathrm{ϕ}}\right)\hat{r}\\ +\frac{1}{r}\left(\frac{1}{\sin \mathrm{θ}}\frac{∂r{\cos }^2\mathrm{θ}}{∂\mathrm{ϕ}}-\frac{∂\left(r\right(3r\left)\right)}{∂r}\right)\widehat{\mathrm{θ}}+\frac{1}{r}\left(\frac{∂\left(r\right(r\cos \mathrm{θ}\sin \mathrm{θ}}{∂r}-\frac{∂(r{\cos }^2\mathrm{θ})}{∂\mathrm{θ}}\right)\widehat{\mathrm{ϕ}}\end{array}\right]\begin{array}{}\end{array}$

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

The differential elemental area is $da=rdrd\mathrm{θ}\widehat{\mathrm{ϕ}}$ . Substitute role="math" localid="1654664744003" $3cot\hat{r}-\hat{6}\mathrm{θ}$for $∇×v$, into the stokes theorem $∫(∇×v).d\mathrm{τ}=-∫v.da$

role="math" localid="1654665787372" $$\begin{gathered} ∫(∇×v).d\mathrm{τ}=0+\left(\underset{0}{\overset{1}{∫}}6rdr\underset{0}{\overset{\frac{\mathrm{π}}{2}}{∫}}d\mathrm{ϕ}\right) \\ =6{{\left(\frac{r^2}{2}\right)}^1}_0\left(\mathrm{ϕ}\right)\stackrel{\frac{\mathrm{π}}{2}}{0} \\ =\left(3\right)\left(\frac{\mathrm{π}}{2}\right) \\ =\frac{3\mathrm{π}}{2}...................\left(1\right) \end{gathered}$$

The differential length vector is given by$dl=dr\hat{r}+rd\mathrm{θ}\widehat{\mathrm{θ}}+r\sin \mathrm{θ}\widehat{\mathrm{ϕ}}$ . Along the path, localid="1654672580270" $\mathrm{θ}=\frac{\mathrm{π}}{2},\mathrm{θ}=0$ 0 and r varies from 0 to 1.Hence the line integral becomes,

localid="1654672141711" $$\begin{gathered} ∫\mathrm{v}.\mathrm{dl}=∫\left((\mathrm{r}{\cos }^2\mathrm{θ}\right)\hat{\mathrm{r}}-(\mathrm{r}\mathrm{³¦´Ç²õθ}\mathrm{²õ¾\pm ²Ôθ})\widehat{\mathrm{θ}}+3\mathrm{r}\widehat{\mathrm{Ï•}}\left)\right(\mathrm{dr}\hat{\mathrm{r}}+\mathrm{»åθ}\widehat{\mathrm{θ}}+\mathrm{r}\mathrm{²õ¾\pm ²Ôθ»åÏ•}\widehat{\mathrm{Ï•}}) \\ =∫(\mathrm{r}{\cos }^2\mathrm{θ})\mathrm{dr}-({\mathrm{r}}^2\mathrm{³¦´Ç²õθ}\mathrm{²õ¾\pm ²Ôθ})\mathrm{»åθ}+3{\mathrm{r}}^2\mathrm{²õ¾\pm ²Ôθ»åÏ•} \\ =∫\left(\mathrm{r}{\cos }^2\left(\frac{\mathrm{Ï€}}{2}\right)\right)\mathrm{dr}-\left({\mathrm{r}}^2\cos \left(\frac{\mathrm{Ï€}}{2}\right)\sin 3{\mathrm{r}}^2\sin \left(\frac{\mathrm{Ï€}}{2}\right)\right)\mathrm{»åθ}3{\mathrm{r}}^2\sin \left(\frac{\mathrm{Ï€}}{2}\right)\mathrm{»åÏ•} \\ =∫0\mathrm{dr}-0\mathrm{»åθ}+3{\mathrm{r}}^2\mathrm{»åÏ•} \end{gathered}$$

Simplify further as

localid="1654672150719" $$\begin{gathered} ∫\mathrm{v}.\mathrm{dl}=∫3{\mathrm{r}}^2\mathrm{»åÏ•} \\ =3{\mathrm{r}}^2\mathrm{Ï•} \\ =3{\mathrm{r}}^2\left(0\right) \\ =0 \end{gathered}$$

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

localid="1654672163671" $$\begin{gathered} ∫\mathrm{v}.\mathrm{dl}=∫\left((\mathrm{r}{\cos }^2\mathrm{θ}\right)\hat{\mathrm{r}}-(\mathrm{r}\mathrm{³¦´Ç²õθ}\mathrm{²õ¾\pm ²Ôθ})\widehat{\mathrm{θ}}+3\mathrm{r}\widehat{\mathrm{Ï•}}\left)\right(\mathrm{dr}\hat{\mathrm{r}}+\mathrm{»åθ}\widehat{\mathrm{θ}}+\mathrm{r}\mathrm{²õ¾\pm ²Ôθ»åÏ•}\widehat{\mathrm{Ï•}}) \\ =∫(\mathrm{r}{\cos }^2\mathrm{θ})\mathrm{dr}-({\mathrm{r}}^2\mathrm{³¦´Ç²õθ}\mathrm{²õ¾\pm ²Ôθ})\mathrm{»åθ}+3{\mathrm{r}}^2\mathrm{²õ¾\pm ²Ôθ»åÏ•} \\ =∫(\left(1\right){\cos }^2\left(\frac{\mathrm{Ï€}}{2}\right))\mathrm{dr}-(\left(1^2\right)\cos \left(\frac{\mathrm{Ï€}}{2}\right)\sin \left(\frac{\mathrm{Ï€}}{2}\right)3{\mathrm{r}}^2\sin \left(\frac{\mathrm{Ï€}}{2}\right))\mathrm{»åθ}+3{\mathrm{r}}^2\sin (\frac{\mathrm{Ï€}}{2})\mathrm{»åÏ•} \\ =∫3\mathrm{»åÏ•} \end{gathered}$$

Simplify further as,

$$\begin{gathered} ∫\mathrm{v}.\mathrm{dl}=\underset{0}{\overset{\frac{\mathrm{Ï€}}{2}}{∫}}3\mathrm{»åÏ•} \\ =3\left(\mathrm{Ï•}\right)\stackrel{\frac{\mathrm{Ï€}}{2}}{0} \\ =\frac{3\mathrm{Ï€}}{2} \end{gathered}$$

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

localid="1654672691611" $$\begin{gathered} ∫\mathrm{v}.\mathrm{dl}=∫\left((\mathrm{r}{\cos }^2\mathrm{θ}\right)\hat{\mathrm{r}}-(\mathrm{r}\mathrm{³¦´Ç²õθ}\mathrm{²õ¾\pm ²Ôθ})\widehat{\mathrm{θ}}+3\mathrm{r}\widehat{\mathrm{Ï•}}\left)\right(\mathrm{dr}\hat{\mathrm{r}}+dr\hat{r}+\mathrm{r»åθ}\widehat{\mathrm{θ}}\mathrm{r}\mathrm{²õ¾\pm ²Ôθ»åÏ•}\widehat{\mathrm{Ï•}}) \\ =∫(\mathrm{r}{\cos }^2\mathrm{θ})\mathrm{dr}-({\mathrm{r}}^2\mathrm{³¦´Ç²õθ}\mathrm{²õ¾\pm ²Ôθ})\mathrm{»åθ}+3{\mathrm{r}}^2\mathrm{²õ¾\pm ²Ôθ»åÏ•} \\ =∫(\frac{1}{\sin \mathrm{θ}}{\cos }^2\mathrm{θ}\left)\right(-\frac{1}{\sin \mathrm{θ}}\cos \mathrm{θ}d\mathrm{θ})-({\left(\frac{1}{\sin \mathrm{θ}}\right)}^2\cos \mathrm{θ}\sin \mathrm{θ})d\mathrm{θ}+0 \end{gathered}$$

Simplify further as,

localid="1654670347504" $$\begin{gathered} ∫\mathrm{v}.\mathrm{dl}=-∫\left(\frac{{\cos }^3\mathrm{θ}}{{\sin }^3\mathrm{θ}}+\frac{\mathrm{³¦´Ç²õθ}}{\mathrm{²õ¾\pm ²Ôθ}}\right)\mathrm{»åθ} \\ =∫\frac{\mathrm{³¦´Ç²õθ}}{\mathrm{²õ¾\pm ²Ôθ}}\left(\frac{{\cos }^2\mathrm{θ}+{\sin }^2\mathrm{θ}}{{\sin }^2\mathrm{θ}}\right)\mathrm{»åθ} \\ =\underset{\frac{\mathrm{Ï€}}{2}}{\overset{{\tan }^{-1}{\left(\frac{1}{2}\right)}^2}{∫}}\frac{\mathrm{³¦´Ç²õθ}}{{\sin }^3\mathrm{θ}}\mathrm{»åθ}...........\left(2\right) \end{gathered}$$

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

$$\begin{gathered} ∫v.dl=-∫v.dl=\frac{1}{2{\sin }^2\mathrm{θ}} \\ =\frac{1}{2x^2} \end{gathered}$$

Substitute back for into above result as,

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

Evaluate the limit as,

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

Along the path (iv), ${\text{θ=³Ù²¹²Ô}}^{-1}{\left(\frac{1}{2}\right)}_{,}\mathrm{Ï•}=\frac{\mathrm{Ï€}}{2}$and r varies from localid="1654671277385" $\sqrt{5}$ to 0, Hence the line integral becomes,

localid="1654672112340" $$\begin{gathered} ∫\mathrm{v}.\mathrm{dl}=∫\left(\left({\mathrm{rcos}}^2\mathrm{θ}\right)\right)\hat{\mathrm{r}}-(\mathrm{r}\mathrm{³¦´Ç²õθ²õ¾\pm ²Ôθ})\widehat{\mathrm{θ}}+3\mathrm{r}\widehat{\mathrm{Ï•}}\left)\right(\mathrm{dr}\hat{\mathrm{r}}+\mathrm{r»åθ}\widehat{\mathrm{θ}}+\mathrm{r²õ¾\pm ²Ôθ»åÏ•}\widehat{\mathrm{Ï•}} \\ =∫\left(\mathrm{r}{\cos }^2\left({\tan }^{-1}\left(\frac{1}{2}\right)\right)\right)\mathrm{dr} \\ =\underset{\sqrt{5}}{\overset{0}{∫}}0.8\mathrm{rdr} \\ =0.8{{\left(\frac{{\mathrm{r}}^2}{2}\right)}^0}_{\sqrt{5}} \end{gathered}$$

Simplify further as,

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

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

$$\begin{gathered} âˆ\circledR v.dl=0+\frac{3\mathrm{Ï€}}{2}+2+\left(-2\right) \\ =\frac{3\mathrm{Ï€}}{2}.........\left(3\right) \end{gathered}$$

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