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 .


2 Answers 2


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.


It is (i) that is correct, not (ii). This is because it is part of the surface of the sphere. It is constrained to lie inside the cyllinder, but this doesn't mean that the surface is part of the cyllinder's surface.

Its not quite right to say $S = \nabla S$. What it is, is that whenever you have a surface defined as a level set of some function, say $G(x,y,z) = c$ defines the surface $S$, then a normal vector of the surface $S$ is parallel to $\nabla G$. (Don't forget to normalise the vector if you need to)

Also, for a sphere centered at 0, it is a cute fact (which is geometrically intuitive, imo) that the position vector is normal to the surface.

I also encourage you to do (iii) because practice makes you less error-prone.

  • $\begingroup$ Thanks, for the answer. I have just one more doubt, Suppose I need to find a Line integral $\int y \hat{i} + z\hat{j} + x\hat{k} .dr$ corresponding to the curve which is intersection of $x^2 + y^2 + z^2 = 1$ and $ x + y = 0$ then for calculating normal vector I can simply calculate $\nabla (x^2 + y^2+ z^2 -1)$. will this be correct ? $\endgroup$
    – zeroflank
    Oct 26, 2019 at 7:54
  • $\begingroup$ @zeroflank You're welcome, yes, but you need the normal of $x+y=0$ also, then take the cross product, this gives you a vector tangent to the intersection curve. $\endgroup$ Oct 26, 2019 at 7:59
  • $\begingroup$ Oh ,Actually I want to solve the previous question with help of Stoke's theorem then is My normal vector correct ? $\endgroup$
    – zeroflank
    Oct 26, 2019 at 8:02
  • $\begingroup$ Oh I see, yes that works $\endgroup$ Oct 26, 2019 at 8:03
  • $\begingroup$ Thank You very much, You have been a great help! $\endgroup$
    – zeroflank
    Oct 26, 2019 at 8:08

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