Applying stokes' theorem to evaluate the double integral for the hemisphere with radius 2 for $y \geq 0$ and oriented in the $+y$ direction. A question on my homework asks:
Use Stokes' Theorem to evaluate $\int\int_{S} (curl\vec{F}) * {n} \space dS$
where S is defined as the hemisphere $x^2 + y^2 + z^2 = 4$ for $y \geq 0$ and oriented in the positive y direction. 
They've also given the vector field $\vec{F} = \space <ze^y, xcos(y), xzsin(y) > $  
So I started by drawing the surface, which is pretty straightforward. I'm gonna be integrating over the xz plane, where the curve is defined by $x^2 + z^2 = 4$, i.e. a circle with radius of $2$. According to my book, the orientation of the curve is consistent with that of the surface if you imagine a person walking around the curve and 1) their head points in the same direction as the normal vector to the surface, and 2) the surface is to their left at each point along the curve.
I parametrized the curve as follows:
$\vec{r(t)} = \space <2cos(t), 0, 2sin(t)>$
I did everything else right--I plugged in the values for x, y, and z into the vector field and properly set up the integral. In the end, I got an answer of $-4\pi$, but this was wrong. I then tried entering $4\pi$, and that was apparently the correct answer. This seems to imply that while I did everything else right, I had to set up the limits as $t$ going from $2\pi$ to $0$ as opposed to $0$ to $2\pi$.
But this doesn't make sense to me, since that forces the surface to be to the right of the imaginary person as he walks along the curve.
What did I do wrong?
 A: Your error comes from the fact in the $xz$ plane:
\begin{cases}
x=2\sin t \\
z=2 \cos t
\end{cases}

Note that you can also compute the integral with very little computations as follows:
By the divergence theorem, the integral equals
$$
\iiint_E \underbrace{div(curl\,\vec{F}\,)}_{=0}\; dV - \iint_{S_1}curl\,\vec{F}\,\cdot d\vec{S}
$$
where $S_1$ is the part of the plane $y=0$ that closes the hemisphere, namely the disc $x^2+z^2\le 4$, that you can parametrize as follows:
\begin{cases}
x=x\\
y=0 \quad \text{with}\quad (x,z)\in \{(r,\theta)\mid 0\le \theta \le 2\pi, 0\le r \le 2 \}\\
z=z
\end{cases}
So the integral equals
$$
-\iint_{D}\pmatrix{xz\cos 0 \\e^{0}-z\sin 0\\\cos 0 - z e^{0}}\cdot \pmatrix{0\\-1\\0}\; dA = \iint_{D} 1\; dA = 4 \pi
$$
A: The curve you've chosen is parametrized clockwise: as the person walks, the surface is at the right. This is why when you interchange the integration limits you get the right sign: you are following the path in the reverse, so is, with the right orientation. I think you simply used the usual parametrization of the circle when in the $xy$ plane, but it doesn't work because the $x$ axis is "reversed".
This one will do the job:
$$\vec{r(t)} = \space <2cos(t), 0, -2sin(t)>$$
