Evaluating these two integrals Let A be defined as:
$$A=\{(x,y,z)\in \text{unit sphere}|z\ge\cos(\alpha)\} \space0\le\alpha\le\pi$$
Take $\hat N$ as the outer unit normal vector for the sphere, $\hat n$ as the unit vector at the boundary of A which is tangent to the unit sphere, orthogonal to the boundary and points away from A (i.e. $\hat n \cdot \hat N=0)$. Calculate directly (without Stokes' theorem):
$$\int_{bdA}\hat nds$$
and
$$\iint_A\hat Nd\text{(area of A)}$$ 
I have proved that the first is equal to half the second in general, but I'm lost on the specific computation.
 A: Let $(r,\theta,\phi)$ be spherical coordinates such that the surface of the unit sphere is given by $r=1$.  Then, the set $A$ is the set for which $r=1$, $\theta \ge \alpha$, and $0\le \phi < 2\pi$.  
The boundary, $\partial A$, of $A$ is the circle $r=1$, $\theta=\alpha$, and $0\le \phi<2\pi$.  
The unit tangent to that circle is $\hat \phi=-\hat x\sin(\phi)+\hat y\cos(\phi)$.  T
The unit vector that is tangent to the sphere, and normal to $\partial A$ is the unit vector 
$$\begin{align}\hat n(\alpha,\phi)&= \hat \theta(\alpha,\phi)\\\\
& =\hat x \cos(\alpha)\cos(\phi)+\hat y \cos(\alpha)\sin(\phi)-\hat z \sin(\alpha)
\end{align}$$
The integral of $\hat n(\alpha,\theta)$ over $\partial A$ is given by
$$\begin{align}
\oint_{\partial A}\hat n\,d\ell&=\int_0^{2\pi} (\hat x \cos(\alpha)\cos(\phi)+\hat y \cos(\alpha)\sin(\phi)-\hat z \sin(\alpha))\,\sin(\alpha)\,d\phi\\\\
&=-\hat z 2\pi \sin^2(\alpha)
\end{align}$$
where we exploited the facts that $\int_0^{2\pi}\sin(\phi)\,d\phi=\int_0^{2\pi}\cos(\phi)\,d\phi=0$.
The integral of $\hat N(\theta,\phi)=\hat r(\theta,\phi)$ over $A$ is
$$\begin{align}
\int_A \hat N dS&=\int_0^{2\pi}\int_{0}^{\alpha}(\hat x\sin(\theta)\cos(\phi)+\hat y\sin(\theta)\sin(\phi)+\hat z\cos(\theta))\sin(\theta)\,d\theta\,d\phi\\\\
&=\hat z \pi \sin^2(\alpha)
\end{align}$$
where we again exploited the facts that $\int_0^{2\pi}\sin(\phi)\,d\phi=\int_0^{2\pi}\cos(\phi)\,d\phi=0$ and also used $\int_0^\alpha \sin(\theta)\cos(\theta)\,d\theta=\frac12 \sin^2(\alpha)$.
