Verify stokes theorem in spherical coordinates

Question

I got the vector field $F= r \sin^2 (\theta) \hat{r} + r\sin(\theta)\cos(\phi) \hat{\theta}+ r\cos^2(\theta)\sin(\phi) \hat{\phi}$. I’m trying to solve the integral $ \int_C F \cdot dl$ over a close path. The idea is to use the volume stored in the upper hemisphere of radius $R$ $(x^2 + y^2 + z^2 =R^2, z>0)$ but I cannot use the stokes theorem, since the idea is to verify both integrals have the same result. I don’t know how to parametrize the surface. I know $dl= dr \hat{r} + rd\theta \hat{\theta} + r\sin(\theta) d\phi \hat{\phi}$ but I cannot integrate over a closed path with this. Thank you!

Answer 1

Parameterizing an upper hemisphere in spherical coordinates is no problem at all...

Stokes's Theorem states;

$$\int_{S}({\nabla \times F})\cdot dS=\oint_{\partial S}F\cdot ds$$

So, for our case we have;

$$\int_{S}[({\nabla \times \vec F})\cdot{\hat r}] dA=R\int_{0}^{2\pi}[\vec F\cdot {\hat\theta}] d\theta$$

$$\int_{S}[({\nabla \times \vec F})\cdot{\hat r}] dA=R\int_{0}^{2\pi}[R\sin(\theta)\cos({\pi\over2})] d\theta$$

$$\therefore \int_{S}[({\nabla \times \vec F})\cdot{\hat r}] dA=0$$

The formula for curl in spherical coordinates is a bit cumbersome but eventually you will arrive at...

$$\int_{S}[({\nabla \times \vec F})\cdot{\hat r}] dA$$

$$=R^2\left(\int_{0}^{\pi}(\cos^2(\phi)-\sin^2(\phi))d\phi\int_{0}^{2\pi}\sin(\theta)d\theta \space-\space \int_{0}^{\pi}\sin(\phi)d\phi\int_{0}^{2\pi}\cos(\theta)\sin(\theta)d\theta\right)$$

$$=0$$

So, as expected, we have;

$$\int_{S}({\nabla \times F})\cdot dS=\oint_{\partial S}F\cdot ds=0$$