First, we check the line integral. The contour is parameterized by
\[
\gamma(\theta) = [a\cos\theta, a\sin\theta, 0],
\]

for $\theta \in [0, 2\pi]$. We then have
\[
\int_C \mb{F} \, \de{\mb{r}} = \int_0^{2\pi} \mb{F} \cdot \gamma'(\theta) \, \de{\theta} = a^2\pi.
\]

Next, we compute the surface integral portion of Stokes' Theorem. We can verify that
\[
\nabla \wedge \mb{F} = [0, 0, 2x + 1],
\]

so we must integrate
\[
\iint_S (2x + 1)\mb{k} \cdot \mb{n} \, \de{S}.
\]

Now the unit normal of the surface is given by $\mb{n} = [x, y, z]/a$. We will then transform to spherical coordinates, and use the Jacobian to write $\de{S} = a^2 \sin\phi \, \de{\phi}\de{\theta}$ for azimuthal coordinate $\phi \in [0, \pi]$. This leaves us with
\[
\int_0^{2\pi} \int_0^{\pi/2} (2r\cos\theta\sin\phi + 1) \left(\frac{a\cos\phi}{a}\right) (a^2 \sin\phi) \, \de{\phi} \de{\theta} = a^2\pi,
\]

which turns out is zero. So that verifies Stokes' Theorem. Notice that instead of arguing through the components (i.e. by constructing the normal vector through intuition), you could do things more systematically by setting $\mb{r}(\theta, \phi)$ to be the position vector on the surface, and compute the usual cross product. But you would find that you are recomputing the Jacobian and the same result would be obtained (this is good practice! do it!)