So, here is the question :
Find $\displaystyle \int\int_{S} \text{curl F}ds$ where $\vec{F} = xz\hat{i} + yz\hat{j} + xy\hat{k}$ and $S$ is the part of the sphere $x^2 + y^2 + z^2 =1$ that lies inside the cylinder $x^2 + y^2 =1$ above $x-y$ plane.
My main issue is such question is how can I compute the outward Normal Vector ?
These are the following things that come to my mind:
Since Normal Vector for a Surface $S$ = $\nabla S$ so we can calculate
(i) $\nabla (x^2 + y^2 +z^2 -1)$
But since we are working on the surface inside the cylinder any Vector Normal to this cylinder should do the job.
So we can calculate
(ii) $\nabla(x^2 + y^2 -1)$
The third method could be to define $x = rcos\theta$ ,$y = rsin\theta$ and $z =\sqrt{3}$
and define $\vec{f} = rcos\theta\hat{i} + rsin\theta\hat{j} + \sqrt{3}\hat{k}$
and then
(iii) we calculate $\dfrac{\partial{F}}{\partial{r}}\times \dfrac{\partial{F}}{\partial{\theta}}$
These are the three method that come to my mind , I would like to avoid (iii) because it is very lengthy and error prone.
But I am really confused between (i) and (ii) , Which one among them is correct and why ? or are they both correct ?
Can someone please answer these doubts ?
Thank you .
Best Answer
The outward normal vector for the sphere is
$$\vec n =\frac1{\sqrt{x^2+y^2+z^2}}\left(x,y,z\right)$$
Note that the sphere and the cylinder have in common the circle in the $x-y$ plane therefore $S$ should be the hemisphere over the $x-y$ plane.