I have a question about Stokes Theorem. It's really confusing me.
So the problem asks,
Use Stokes theorem to set up the integral $$\int_{a}^{b}g(t)dt$$ that you would use to compute $$\iint_{S}^{}(\nabla\times \vec{F})\cdot d\vec{S}$$ where $$\vec{F}(x,y,z)=\left \langle e^zsin(y),ze^x,2xcos(y) \right \rangle$$
and S is the hemisphere $y=\sqrt{9-x^2-z^2}$ oriented in the direction of the positive y-axis.
What I did was first calculate the curl:
$$\nabla\times \vec{F}=\left \langle -2xsin(y)-e^zcos(y),e^zsin(y)-2cos(y),ze^x-e^zcos(y) \right \rangle$$
Skipping steps,I have $$\frac{\partial \vec{r}}{\partial x}\times \frac{\partial \vec{r}}{\partial z} =\left \langle -\frac{x}{\sqrt{9-x^2-z^2}},-1,-\frac{z}{\sqrt{9-x^2-z^2}} \right \rangle=\vec{n}$$
$$\hat{n}=\left \langle -\frac{x}{3},-\frac{\sqrt{9-x^2-z^2}}{3},-\frac{z}{3} \right \rangle$$
I'm not sure where to go from here.
As per Stokes' Theorem,
$\int_C \vec{F} \cdot d\vec{r} = \iint_S curl \vec{F} \cdot d\vec{S}$
which allows you to change the surface integral of the curl of the vector field to the line integral of the vector field around the boundary of the surface. The surface is hemisphere with $y = 0$ plane being the boundary, though the question should have been more clear on that. Also the question states we have orientation in positive $y$ direction.
By the way, there is a hint in the question to do line integral to find surface integral of the curl of the vector field.
Now we can parametrize the boundary as,
$\vec{r}(t) = (3 \cos t, 0, 3 \sin t) \,$ (as $y = 0$, it is a circle in $XZ$ plane with radius $3,$ $x^2 + z^2 = 9$)
$\vec{r'}(t) = (-3 \sin t, 0, 3 \cos t) \, $ ($0 \leq t \leq 2 \pi)$
Now substitute $x, y, z$ from above parametrization of the curve to evaluate $\vec{F}(r(t))$ and do the dot product with $\vec{r'}(t)$. Then do line integral wrt $dt$.