Compute the line integral of
v=\left(r\cos ^2\theta \right) \hat{r}-\left(r\cos \theta \sin \theta \right)\hat{\pmb{\theta} }+3r\hat{\pmb{\phi} }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: 3π/2]
Start at the origin
(1) \theta =\frac{\pi }{2} , \phi= 0; r : 0 \rightarrow 1. v · dl =\left(r\cos ^2\theta \right)(dr) = 0.(dr) = 0. \int{}v · dl = 0 .
(2) r = 1, \theta =\frac{\pi }{2}; \phi :0\rightarrow \pi /2. v · dl = (3r)\left(r\sin \theta d\phi \right)=3d\phi . \int{} v · dl = 3\int\limits_{0}^{\pi /2}{}d\phi =\frac{3\pi }{2} .
(3) \phi =\frac{\pi }{2}; r\sin \theta = y = 1, so r =\frac{1}{\sin \theta }, dr =\frac{-1}{\sin ^2\theta } \cos \theta d\theta , \theta : \frac{\pi }{2} \rightarrow \theta _0\equiv \tan ^{-1}\left(1/2\right) .
v · dl =\left(r\cos ^2\theta \right)(dr) −\left(r\cos \theta \sin \theta \right)\left(rd\theta \right)=\frac{\cos ^2\theta }{\sin \theta } \left(-\frac{\cos \theta }{\sin ^2\theta } \right)d\theta -\frac{\cos \theta \sin \theta }{\sin ^2\theta }d\theta=-\left(\frac{\cos ^3\theta }{\sin ^3\theta }+\frac{\cos \theta }{\sin \theta } \right)d\theta =-\frac{\cos \theta }{\sin \theta }\left(\frac{\cos ^2\theta +\sin ^2\theta }{\sin ^2\theta } \right)d\theta =-\frac{\cos \theta }{\sin ^3\theta }d\theta .
Therefore
\int{} v · dl = −\int\limits_{\pi /2}^{\theta _0}{} \frac{\cos \theta }{\sin ^3\theta }d\theta =\frac{1}{2\sin ^2\theta }\mid ^{\theta _0}_{\pi /2}=\frac{1}{2\cdot \left(1/5\right) }-\frac{1}{2\cdot \left(1\right) }=\frac{5}{2}-\frac{1}{2}=2 .
(4) \theta =\theta _0, \phi =\frac{\pi }{2}; r :\sqrt{5}\rightarrow 0. v · dl =\left(r\cos ^2\theta \right)(dr) =\frac{4}{5}r dr .
\int{}v · dl =\frac{4}{5} \int\limits_{\sqrt{5} }^{0}{} r dr =\frac{4}{5}\frac{r^2}{2} \mid ^{0 }_{\sqrt{5} } =-\frac{4}{5}\cdot \frac{5}{2}=-2 .
Total:
\oint{}v · dl = 0+\frac{3\pi }{2} +2-2=\frac{3\pi }{2} .
Stokes’ theorem says this should equal ∫(∇×v) · da
∇×v=\frac{1}{r\sin \theta }\left[\frac{\partial}{\partial\theta } \left(\sin \theta 3r\right) -\frac{\partial}{\partial \phi }\left(-r\sin \theta \cos \theta \right) \right]\hat{r} +\frac{1}{r} \left[\frac{1}{\sin \theta }\frac{\partial}{\partial \phi } \left(r\cos ^2\theta \right)-\frac{\partial}{\partial r} (r3r) \right]\hat{\pmb{\theta} }+\frac{1}{r} \left[\frac{\partial}{\partial r} \left(-rr\cos \theta \sin \theta \right)-\frac{\partial}{\partial \theta }\left(r\cos ^2\theta \right) \right] \hat{\pmb{\phi} }
=\frac{1}{r\sin \theta } \left[3r\cos \theta \right] \hat{r} +\frac{1}{r} \left[-6r\right] \hat{\pmb{\theta} } +\frac{1}{r}\left[-2r\cos \theta \sin \theta +2r\cos \theta \sin \theta \right] \hat{\pmb{\phi} }
=3\cos \theta \hat{r} -6\hat{\pmb{\theta} } .
(1) Back face: da = −r dr d\theta \hat{\pmb{\phi} } ; (∇\times v) · da = 0. \int{}\left(∇\times v\right) · da = 0 .
(2) Bottom: da = −r \sin \theta dr d\phi \hat{\pmb{\theta} }; \left(∇\times v\right)· da = 6r\sin \theta dr d\phi . \theta =\frac{\pi }{2}, so \left(∇\times v \right) · da = 6r dr d\phi
\int{} \left(∇\times v \right)· da =\int\limits_{0}^{1}{}6r dr\int\limits_{0}^{\pi /2}{}d\phi =6\cdot \frac{1}{2}\cdot \frac{\pi }{2}=\frac{3\pi }{2} .