How can I solve this integral with help Stokes formula:
$$\oint_L (y-z)\,dx + (z-x)\,dy + (x-y)\,dz,$$
where $L$ is the circle: $x^2 + y^2 + z^2 = R^2,\; y=x\tan(\varphi),\; \varphi \in \left(0, \pi \right)$ oriented negatively with respect to the vector $(1, 0, 0).$
I think $y=x\tan(\varphi)$ cuts off a circle with radius $1$ and center in $(0,0)$ point from the sphere. But then I am stuck, because I don't know what surface I need. Maybe hemisphere or any plane with any tilt angle.
Sorry, if this task a very simple, but only today I got acquainted with this formula. Thank you in advance!
We have a circle:
$ L = \begin{cases} x^2 + y^2 + z^2 = R^2\\ y = xtan(\phi) \end{cases} $ where $ \phi \in (0, \pi)$
Our integral transform in surface integral by the Stokes formula:
$ \iint_{S} det\begin{pmatrix} cos\alpha & cos\beta & cos\gamma\\ \frac{\partial }{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{pmatrix} = -2(cos\alpha + cos\beta + cos\gamma)\cdot mes\left| {S} \right| = -2\pi r^2(cos\alpha + cos\beta + cos\gamma) $
Cosines uphere is const, because $ S = \begin{cases} x^2 + y^2 + z^2 \leq R^2\\ y = xtan(\phi) \end{cases} $ where $ \phi \in (0, \pi)$, is a plane.
Find a normal vector to determine the cosines.
$\overline {\nu} = \pm \frac{(A, B, C)}{\sqrt {A^2 + B^2 + C^2}} = \pm \frac{(1-tan\phi, 1, 0)}{\sqrt {(tan\phi)^2 + 1}} = \pm (-\frac{sin\phi\left| {cos\phi} \right|}{cos\phi}, \left| {cos\phi} \right|, 0) = \pm (-sin\phi sgn(cos\phi), \left| {cos\phi} \right|, 0) $
Then for $\phi \in (0, \frac{\pi}{2})$ we have the answer: $ 2\sqrt{2} \cdot sin(\frac{\pi}{4} - \phi) \pi R^2 $
But for $ \phi \in (\frac{\pi}{2}, \pi)$ a little harder. There are a normal vectors change a direction. Consequently we have our answer finally:
$ \oint_L (y-z)\,dx + (z-x)\,dy + (x-y)\,dz $ = $ 2\sqrt{2} \cdot sin(\frac{\pi}{4} - \phi) \pi R^2 $