We wish to verify Stokes theorem for a Cone: $z=\sqrt{x^2+y^2}$, $z\leq 3$ for ${\bf F}=\left [y , -x ,2z \ \right ]$
Curl: $\nabla \times {\bf F} =[0,0,-2] $.
$\int\int_S(\nabla \times {\bf F})\cdot{\bf n} dS$
This surface integral for the side of Cone: $ =\int_0^{2\pi} \int_0^3 [0,0,-2] [r \cos \theta, r \sin \theta , -r] dr d\theta$ (Length of Unity Normal vektor is Equal Jacobi and they have been eliminated) $=\int_0^{2\pi} \int_0^3 2r dr d\theta = 2\pi \left [r^2\right ]_0^3= 18 \pi$
Surface integral for the topp of Cone:
$\int \int_{S}(\nabla \times {\bf F})\cdot{\bf n} dS =\int_0^{2\pi} \int_0^3 [0,0,-2] [0, 0, 1] r dr d\theta$
$=\int_0^{2\pi} \int_0^3 -2 r dr d\theta = 2\pi \left [-r^2\right]_0^3= -18 \pi$
Linjeintegral is:
$ \int_C {{\bf F} ({\bf r}(t)}\cdot{\bf r}'(t) dt $ \vskip.2cm $=\int_0^{2\pi} [ r \sin \theta , - r \cos \theta ,2 \cdot 3 ] [-r\sin \theta, r \cos \theta, 0] d \theta = \int_0^{2\pi} -r^2 \ d \theta = \int_0^{2\pi} -9 d \theta =-18 \pi$
Why the Surface integral on the sides has opposite sign? is $-18 \pi$ correct or $18 \pi$? why? Once I heard, we should adjust the sign of uniy normal vector, but this vector should directed to outside of Cone and in this calculation should be correct.
You have the line integral of ${\bf F}$ along $C$, whereby $C$ is oriented such that it is counterclockwise when viewed from high above. The result is $-18\pi$. Correct.
The chosen orientation of $C$ means that the positive normal on the top surface $S_{\rm top}$ of the cone is upwards, i.e., ${\bf n}=(0,0,1)$. You have chosen this and obtained $-18\pi$ for the surface integral of ${\rm curl}({\bf F})$. Correct.
Now you test the Stokes' theorem a second time with $C$ bounding the cone surface $S_{\rm cone}$. Now the direction of $C$ has to make an ordinary screw with the ${\bf n}$ on $S_{\rm cone}$, and this is again the case when ${\bf n}$ points upwards (think about it, looking upwards from down below). But you have chosen downwards. False.
Note that "inner" and "outer" play no role here. The interior of the cone plays no role in this game. You have just replaced the line integral along $C$ by two different surface integrals, whereby both surfaces $S_i$ satisfy $C=\partial S_i$.