I have a region R = {$(x,y,z): x^2 + y^2 + z^2 =1, cosB \leq z \leq cos A$ }, where $ 0 \lt A \lt B \lt \pi.$
I am given a vector field in cylindrical polar coordiantes: $\vec F$ = f(z) $\vec e_{\theta}$ and need to verify Stokes' theorem for this surface. However, I end up with two answers which differ by a minus sign and I am unsure of which, if any, is correct.
Surface integral : I parametrise the surface by $\vec r (\theta, z)$ = $((1-z^2)^\frac 12$ $ cos\theta$, $(1-z^2)^\frac 12 sin\theta, z$)
Then $d\vec S = \vec r_{\theta}$ $ \times$ $\vec r_{z}$ = $((1-z^2)^\frac 12$ $ cos\theta$, $(1-z^2)^\frac 12 sin\theta, z$). I can also write $\vec e_{\theta}$ = $(-sin\theta, cos\theta, 0)$, so I get $\nabla \times \vec F$ = $(-f'(z)cos\theta, -f'(z) sin\theta, f'(z)(1-z^2)^\frac {-1}2) $
Then $\iint \nabla \times \vec F \cdot d\vec S $ = $\int_{0}^{2\pi} \int_{cosB}^{cosA} -f'(z)(1-z^2)^\frac 12 + f(z)z(1-z^2)^\frac {-1}2 dzd\theta$
=$\int_{0}^{2\pi} \int_{cosB}^{cosA} \frac d{dz}[-f(z)(1-z^2)^\frac 12 ]dzd\theta$ = $2\pi[f(cosB)sinB-f(cosA)sinA].$
Line integral: The boundary is made up of 2 curves. The top curve is a circle of radius sin(A). To orient this correctly with outward normal, I need to parametrise the curve counterclockwise. Hence $\vec r(\theta) = (sinA cos\theta, cosA sin\theta, cosA)$ is my parametrisation of the top boundary.
$\int_0^{2\pi} \vec F(\vec r(\theta)) \cdot \frac{d\vec r}{d\theta} d\theta$ = $\int_0^{2\pi} (-f(cosA)sin\theta, f(cosA) cos\theta, 0) \cdot (-sinAsin\theta, sinAcos\theta, 0) d\theta$ = $2\pi f(cos(A))sin(A)$.
I do the same for the lower curve and sum these line integrals.
However, my line integral is minus my surface integral. I'm not sure if my orientation is incorrect or if my method is wrong to begin with.
If you are orienting the surface with the normal vector that is facing outwards of the sphere, by the right hand rule the induced orientation on the upper circle is clockwise (seen from above) and on the lower circle is counterclockwise. Picture attached.
This means your surface integral is ok but your line integral has the wrong sign.