How can I calculate the integral $$\iint_S (\nabla \times F)\cdot dS$$ where $S$ is the part of the surface of the sphere $x^2+y^2+z^2=1$ and $x+y+z\ge 1$, $F=(y-z, z-x, x-y)$.
I calculated that $\nabla\times F=(-2 , -2 ,-2)$. It's difficult for me to find the section between the sphere and the plane. Also, I can't calculate the integral.
Update: I found that the result is: $-6π/3^{1/2}$. Ιs it correct?
The result $-6\pi /\sqrt{3}$ appears to be close, but not quite correct. I'll recheck using the approach from comment:
The plane has unit normal vector $(1/\sqrt{3},1/\sqrt{3},1/\sqrt{3})$. The dot product with the field $\nabla \times F$ is $-2\sqrt{3}$. So, the flux is $-2\sqrt{3}$ times the area of the part of the plane within the sphere. This part is a disk of radius... let's see.
On the plane $x+y+z=1 $, the point $A=(1/3,1/3,1/3)$ is orthogonal to the vector $OA$. Hence, the Pythagorean theorem implies that the disk is centered at A and has radius $\sqrt{1-|OA|^2}= \sqrt{1-1/3} = \sqrt{2/3}$. So its area is $2\pi/3$.
Conclusion: the flux is $-4\pi/\sqrt{3}$.
It would be $-6\pi /\sqrt{3}$ if the plane was $x+y+z=0$, for then its part within the sphere would have radius $1$.