Here S is upper half of the sphere x² + y² + z² = 1, i.e., the portion of sphere above the plane z = 0 (xy plane).
So the surface C is x² + y² = 1.
For surface C, let,
- [tex]\rm{x=\cos\theta,\quad dx=-\sin\theta\ d\theta}[/tex]
- [tex]\rm{y=\sin\theta,\quad dy=\cos\theta\ d\theta}[/tex]
- [tex]\rm{z=0,\quad dz=0}[/tex]
where θ ∈ [0, 2π].
Stokes' Theorem is as follows.
[tex]\large\displaystyle\boxed{\oint\limits_{\rm{C}}\vec{\rm{F}}\cdot\vec{\rm{dr}}=\iint\limits_{\rm{S}}\left(\vec{\nabla}\times\vec{\rm{F}}\right)\cdot\rm{\hat n\ ds}}[/tex]
We need to verify this theorem.
First consider LHS.
[tex]\small\text{$\displaystyle\longrightarrow\oint\limits_{\rm{C}}\vec{\rm{F}}\cdot\vec{\rm{dr}}=\oint\limits_{\rm{C}}\left(\rm{y\,\hat i+z\,\hat j+x\,\hat k}\right)\cdot\left(\rm{dx\,\hat i+dy\,\hat j+dz\,\hat k}\right)$}[/tex]
[tex]\displaystyle\longrightarrow\oint\limits_{\rm{C}}\vec{\rm{F}}\cdot\vec{\rm{dr}}=\oint\limits_{\rm{C}}\rm{(y\ dx+z\ dy+x\ dz)}[/tex]
Substituting for each term,
[tex]\small\text{$\displaystyle\longrightarrow\oint\limits_{\rm{C}}\vec{\rm{F}}\cdot\vec{\rm{dr}}=\int\limits_{\rm{0}}^{\rm{2\pi}}\rm{\big(\sin\theta(-\sin\theta\ d\theta)+0(\cos\theta\ d\theta)+\cos\theta(0)\big)}$}[/tex]
[tex]\displaystyle\longrightarrow\oint\limits_{\rm{C}}\vec{\rm{F}}\cdot\vec{\rm{dr}}=\int\limits_{\rm{0}}^{\rm{2\pi}}\rm{-\sin^2\theta\ d\theta}[/tex]
Solve this definite integral to get,
[tex]\displaystyle\longrightarrow\oint\limits_{\rm{C}}\vec{\rm{F}}\cdot\vec{\rm{dr}}=-\pi[/tex]
Now we need to check whether RHS has the same value.
For surface S, let,
- [tex]\rm{x=\sin\theta\cos\phi}[/tex]
- [tex]\rm{y=\sin\theta\sin\phi}[/tex]
- [tex]\rm{z=\cos\theta}[/tex]
[tex]\longrightarrow\vec{\nabla}\times\vec{\rm{F}}=\left|\begin{array}{ccc}\rm{\hat i}&\rm{\hat j}&\rm{\hat k}\\\\\rm{\dfrac{\partial}{\partial x}}&\rm{\dfrac{\partial}{\partial y}}&\rm{\dfrac{\partial}{\partial z}}\\\\\rm{y}&\rm{z}&\rm{x}\end{array}\right|[/tex]
[tex]\longrightarrow\vec{\nabla}\times\vec{\rm{F}}=\rm{-(\hat i+\hat j+\hat k)}[/tex]
Define the function [tex]\rm{f(x,\ y,\ z)=x^2+y^2+z^2-1}[/tex] to obtain,
[tex]\longrightarrow\rm{\hat n=\dfrac{\nabla f}{|\nabla f|}}[/tex]
[tex]\small\text{$\longrightarrow\rm{\hat n=\dfrac{\left(\hat i\,\dfrac{\partial}{\partial x}+\hat j\,\dfrac{\partial}{\partial y}+\hat k\,\dfrac{\partial}{\partial z}\right)\big(x^2+y^2+z^2-1\big)}{\left|\left(\hat i\,\dfrac{\partial}{\partial x}+\hat j\,\dfrac{\partial}{\partial y}+\hat k\,\dfrac{\partial}{\partial z}\right)\big(x^2+y^2+z^2-1\big)\right|}}$}[/tex]
[tex]\longrightarrow\rm{\hat n=\dfrac{2(x\,\hat i+y\,\hat j+z\,\hat k)}{2\sqrt{x^2+y^2+z^2}}}[/tex]
Since [tex]\rm{x^2+y^2+z^2=1,}[/tex]
[tex]\longrightarrow\rm{\hat n=x\,\hat i+y\,\hat j+z\,\hat k}[/tex]
And ds is elementary area on the hemisphere given by,
- [tex]\rm{ds=\sin\theta\ d\theta\ d\phi}[/tex]
Now,
[tex]\small\text{$\displaystyle\longrightarrow\iint\limits_{\rm{S}}\left(\vec{\nabla}\times\vec{\rm{F}}\right)\cdot\rm{\hat n\ ds}=-\iint\limits_{\rm{S}}\rm{\left(\hat i+\hat j+\hat k\right)\cdot\left(x\,\hat i+y\,\hat j+z\,\hat k\right)\ ds}$}[/tex]
[tex]\small\text{$\displaystyle\longrightarrow\iint\limits_{\rm{S}}\left(\vec{\nabla}\times\vec{\rm{F}}\right)\cdot\rm{\hat n\ ds}=-\iint\limits_{\rm{S}}\rm{(x+y+z)\ ds}$}[/tex]
Substituting for each term,
[tex]\small\text{$\displaystyle\longrightarrow\iint\limits_{\rm{S}}\left(\vec{\nabla}\times\vec{\rm{F}}\right)\cdot\rm{\hat n\ ds}=-\int\limits_{\rm{\phi=0}}^{\rm{\phi=2\pi}}\,\int\limits_{\rm{\theta=0}}^{\rm{\theta=\frac{\pi}{2}}}\rm{(\sin\theta\cos\phi+\sin\theta\sin\phi+\cos\theta)\sin\theta\ d\theta\ d\phi}$}[/tex]
[tex]\small\text{$\displaystyle\longrightarrow\iint\limits_{\rm{S}}\left(\vec{\nabla}\times\vec{\rm{F}}\right)\cdot\rm{\hat n\ ds}=-\int\limits_{\rm{\phi=0}}^{\rm{\phi=2\pi}}\,\int\limits_{\rm{\theta=0}}^{\rm{\theta=\frac{\pi}{2}}}\rm{(\sin^2\theta\cos\phi+\sin^2\theta\sin\phi+\sin\theta\cos\theta)\ d\theta\ d\phi}$}[/tex]
Solving each integral we get,
- [tex]\displaystyle\rm{\int\limits_{\phi=0}^{\phi=2\pi}\,\int\limits_{\theta=0}^{\theta=\frac{\pi}{2}}\sin^2\theta\cos\phi\ d\theta\ d\phi=0}[/tex]
- [tex]\displaystyle\rm{\int\limits_{\phi=0}^{\phi=2\pi}\,\int\limits_{\theta=0}^{\theta=\frac{\pi}{2}}\sin^2\theta\sin\phi\ d\theta\ d\phi=0}[/tex]
- [tex]\displaystyle\rm{\int\limits_{\phi=0}^{\phi=2\pi}\,\int\limits_{\theta=0}^{\theta=\frac{\pi}{2}}\sin\theta\cos\theta\ d\theta\ d\phi=\pi}[/tex]
Then,
[tex]\displaystyle\longrightarrow\iint\limits_{\rm{S}}\left(\vec{\nabla}\times\vec{\rm{F}}\right)\cdot\rm{\hat n\ ds}=\rm{-(0+0+\pi)=-\pi}[/tex]
So LHS = RHS. Hence Stokes' Theorem is verified!
[tex]\displaystyle\longrightarrow\underline{\underline{\oint\limits_{\rm{C}}\vec{\rm{F}}\cdot\vec{\rm{dr}}=\iint\limits_{\rm{S}}\left(\vec{\nabla}\times\vec{\rm{F}}\right)\cdot\rm{\hat n\ ds}=-\pi}}[/tex]
