$F(x,y,z)= \begin{pmatrix} -y \\ 2x\\z \end{pmatrix}$

be a vector field and $A$ a hemisphere

with $x^2+y^2+z^2=9 $ , $ z>0 $

with a circular edge at the $x,y $- level with the unit normal vector $n$ showing outwards.

I want to determine

$ \int_A ( \nabla \times F) n\; do $

1) as a surface integral
2) with Stoke's theorem.


I used the parametrization $ \Phi(\phi, \theta) =\begin{pmatrix} R\sin \theta \cos\phi \\ R\sin\theta \sin \phi\\R\cos \theta \end{pmatrix}$ with $ 0 \leq \phi \leq 2 \pi $ and $ 0\leq \theta\leq \frac{\pi}{2}$

for the unit normal:

$ \frac{ \delta \Phi }{ \phi} \times \frac{ \delta \Phi}{ \theta} = \begin{pmatrix} R^2\sin^2 \theta \cos \phi \\ R^2\sin^2 \theta \sin \phi \\R^2\sin \theta \cos\theta \end{pmatrix}$

so integrate

$\int_0^{ \frac{ \pi}{2}} \int_0^{2 \pi} \begin{pmatrix} -Rsin\theta \cos \phi \\ 2R\sin \theta \sin \phi \\R \cos\theta \end{pmatrix}\begin{pmatrix} R^2\sin^2 \theta \cos \phi \\ R^2\sin^2 \theta \sin \phi \\R^2\sin \theta \cos\theta \end{pmatrix} d\phi d\theta = 2 \pi R^3 $


for Stoke's theorem I use as parametrization $ \Phi ( \phi) = \begin{pmatrix} -\sin \phi \\ \cos \phi \\ 0 \end{pmatrix} $ because $z=0 .$

Then I get to calculate following:

$\int_0^{2 \pi} \begin{pmatrix} -r \sin \phi \\ 2r\cos \phi \\ -r^2 \end{pmatrix}\begin{pmatrix} -\sin \phi \\ \cos \phi \\ 0 \end{pmatrix} d \phi =\int_0^{2 \pi} r \sin^2 \phi+ 2r \cos^2 \phi d \phi = 3\pi r, $

and with $r=3$ follows $ 9 \pi .$

So, they are not equal. I dont see my mistake. Could not find one in the calculations, so there must be one in the process? I have lost perspective, I appreciate any help a loot !!


1) I don’t understand why didn’t you calculate the curl directly. I got $\nabla \times F = (0,0,3)$. I am sorry but — is the two $R\sin\theta\cos\phi$ a typo?

2) The parametrization part seems incorrect. In the context, the hemisphere has a radius 3. So you should probably go with $ \Phi ( \phi) = \begin{pmatrix} -R\sin \phi \\ R\cos \phi \\ 0 \end{pmatrix} $ instead. Besides, it seems that you didn’t calculate the derivative in the last integration formula.

By the way, in both cases, my answer is $3\times \pi R^2=27\pi$.

EDIT: As a reply to the comment:

This is how I would work out the curve integral. First, find a parametrization such that it moves anti-clockwise as you see from the top. It has to rotate in that way because the Stokes’ theorem involves direction. For example, let $x=R\cos t$, $y=R\sin t$ and $z=z$, where $t\in [0,2\pi]$. Then use $$ \oint_C F(x,y,z) \cdot\mathrm d l= \int_0^{2\pi} F(x,y,z) \cdot (x’(t), y’(t), z’(t)) \mathrm d t$$

Note that we have $\mathrm d l = (x’(t), y’(t), z’(t)) \mathrm d t$. In this case, $$ \oint_C F(x,y,z) \cdot\mathrm d l = \int_0^{2\pi} (-R\sin t, 2R\cos t, 0)\cdot(-R\sin t, R\cos t, 0) \mathrm d x = R^2\int_0^{2\pi} (\sin^2 t + 2\cos^2 t) \mathrm d t = 3\pi R^2 $$

  • $\begingroup$ thank you ! okay, the 1) is clear..i don't know why I didnt see this. But what do you mean at 2) with the derivative? I get $R^2 \pi 3 $ with $ \int_0^{2 \pi} \begin{pmatrix} -Rsin \phi \\2Rcos \phi \\o \end{pmatrix} \begin{pmatrix} -Rsin \phi \\ Rcos \phi \\ 0 \end{pmatrix} $ $\endgroup$
    – constant94
    Dec 12 '18 at 19:58
  • $\begingroup$ @constant94 See my edit. $\endgroup$
    – fantasie
    Dec 13 '18 at 1:11
  • $\begingroup$ Even easier, you can compute the line integral around the circle by applying Stokes's Theorem again, filling in the circle with a disk. So you get $3$ times the area of the circle of radius $R$, namely $3\pi R^2$. $\endgroup$ Dec 13 '18 at 1:35
  • $\begingroup$ That’s true, and the easy way — projecting to the disk — can also be seen from the first method. Here I am just make some comparisons, in order to point out where he is mistaken. $\endgroup$
    – fantasie
    Dec 13 '18 at 1:46

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