I have this pretty simple exercise:
Use Stokes' Theorem to find $\iint_S \text{curl} \, \mathbf{F} \cdot d \mathbf{S}$, where $\mathbf{F}(x,y,z) = ze^y \mathbf{i} + x \cos y \mathbf{j} + xz \sin y \mathbf{k}$, $S$ is the hemisphere $x^2+y^2+z^2 = 16, y \geq 0$, and the orientation is in the direction of the positive $y$-axis.
I got $-16\pi$ but the actual answer is $16\pi$ so I think that the reason for this was not setting up the orientation properly, here is my work:
For the boundary curve $C$ I chose $\mathbf{r}(t) = \langle 4 \cos t, 0, 4\sin t \rangle, t \in [0, 2\pi]$. It follows that $\mathbf{r}'(t) = \langle -4\sin t, 0, 4\cos t \rangle$. From here I will Stokes' Theorem: $$ \iint\limits_S \text{curl} \space \mathbf{F} \cdot d\mathbf{S} = \int_C \mathbf{F} \cdot d\mathbf{r} $$
But $$ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt $$
Plugging in $\mathbf{r}(t)$ into $\mathbf{F}$ gives $$ \mathbf{F}(\mathbf{r}(t)) = \langle 4\sin t \cdot e^0, 4\cos t \cdot \cos 0, 16\cos t \sin t \sin 0 \rangle = \langle 4\sin t, 4\cos t, 0 \rangle $$ And so $$ \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = \langle 4\sin t, 4\cos t, 0 \rangle \cdot \langle -4\sin t, 0, 4\cos t \rangle = -16\sin^2 t $$ Therefore $$ \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt = \int_0^{2\pi} -16\sin^2 t = -16\pi $$
But this is wrong. Can someone show me the right way of doing this? Specifically the orientation part because that's what I'm most struggling with.