Green's theorem in the plane is a special case of Stokes' theorem.
If we express Green's theorem as
$\oint_C M dx + N dy = \iint_R (\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} ) dx dy$
we can express this in vector notation as
$Mdx + Ndy = (M \mathbf{i} + N \mathbf{j})(dx\mathbf{i}+dy\mathbf{j})= \mathbf{A}\cdot d\mathbf{r}$ in which $\mathbf{A} = M\mathbf{i}+ N\mathbf{j}$, $\mathbf{r} = x\mathbf{i}+y\mathbf{j}$.
We have
$\nabla \times \mathbf{A} = -\frac{\partial N}{\partial z}\mathbf{i}+\frac{\partial M}{\partial z}\mathbf{j}+ (\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})\mathbf{k}$
and then $(\nabla \times \mathbf{A})\cdot \mathbf{k} = \frac{\partial N}{\partial x} -\frac{\partial M}{\partial y}$
So now we can re-write Green's theorem as (see * below re "dot" question)
$\oint_C \mathbf{A} \cdot d\mathbf{r} = \iint_R (\nabla x \mathbf{A})\cdot \mathbf{k}$ dR
in which dR = dx dy.
This is basically a problem from Shaum's Vector Analysis (Spiegel)
Edit: the full generalization of
$\oint_C \mathbf{A} \cdot d\mathbf{r} = \iint_R (\nabla x \mathbf{A})\cdot \mathbf{k}$ dR
to the usual version of Stokes' theorem
$\oint_C \mathbf{A} \cdot d\mathbf{r} = \iint_S (\nabla x \mathbf{A})\cdot \mathbf{n}$ dS
is also a problem in Shaum's, which is really an exercise in extending Green's to three dimensions. As Nunoxic points out above, a slightly different treatment of Green's theorem generalizes to Gauss' divergence theorem, also known as Green's theorem in space.
(*) "Dot" question: In your first equation, "curl v" is the result of dotting with the unit normal. I think the usage "curl" is confusing because, comparing your two equations, it appears that you have done something in $R^3$ that you did not do in $R^2$. That is not the case.
What Stokes' Theorem tells you is the relation between the line integral of the vector field over its boundary $\partial S$ to the surface integral of the curl of a vector field over a smooth oriented surface $S$:
$$\oint\limits_{\partial S} \mathbf{F} \cdot d\mathbf{r} = \iint\limits_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} \tag{1}\label{1}$$
Since the prompt asks how to calculate the integral using Stokes' Theorem, you can find a good parametrization of the boundary $\partial S$ and calculate the "easier" integral of the LHS of (1).
Note that the boundary of $S$ is given by:
$$\partial S=\{(x,y,z)\in \mathbb R^3 : x^2 +y^2 =25, z=0\},$$
so a good parametrization to use for $\partial S$ could be:
$$\sigma :[0,2\pi] \subseteq \mathbb{R} \rightarrow \mathbb R^3$$ $$\sigma(\theta)=(5\cos(\theta),5\sin(\theta),0)$$
and finally the integral to calculate ends up being:
$$\begin{align}\oint\limits_{\partial S} \mathbf{F} \cdot d\mathbf{r}&= \iint\limits_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} \\ &=\int_{0}^{2\pi}(0,5\cos(\theta),e^{25\cos(\theta)\sin(\theta)})\cdot(-5\sin(\theta),5\cos(\theta),0)\, d\theta,\\&=\int_{0}^{2\pi}25\cos^2(\theta)\,d\theta\end{align}$$
and from here you can use trigonometric identities to calculate the last integral.
I hope that helps!
Best Answer
Your first calculation is off. We do indeed have $\nabla \times \textbf{F} = \textbf{i}+\textbf{j}+\textbf{k}$ but when we parametrize this "quarter-sphere" as $$\textbf{X}(\phi,\theta)=(\cos \theta \sin \phi , \sin \theta \sin \phi , \cos \phi), \phi \in [0,\pi/2], \theta \in [0,\pi]$$
we have normal vector $\textbf{N}(\phi,\theta) = (\sin^2 \phi \cos \theta, \sin^2 \phi \sin \theta, \cos \phi \sin \phi)$ so evaluating the integral gives us:
$$ \begin{align} \int_{0}^{\pi} \int_{0}^{\pi/2} \left[\sin^2 \phi (\cos \theta + \sin \theta) + \cos \phi \sin \phi \right]\, d\phi \, d\theta &= \frac{1}{4} \int_{0}^{\pi} \left(\pi \sin \theta + \pi \cos \theta + 2 \right)\, d\theta = \pi. \end{align} $$
Now, to use Stoke's theorem, we need a closed boundary so we can parametrize the boundary piecewise as $\textbf{x}_1 \cup \textbf{x}_2$ where
$$ \textbf{x}_1(\theta) = (\cos \theta, \sin \theta, 0) \quad \theta \in [0,\pi] \\ \textbf{x}_2(\theta) = (\cos \theta, 0, \sin \theta) \quad \theta \in [\pi,0]. $$
Evaluating a piecewise line integral gives
$$ \begin{align} \int_{0}^{\pi} (0,\cos \theta,\sin \theta) \cdot (-\sin \theta, \cos \theta, 0) \, d\theta \, + \\ \int_{\pi}^{0} (\sin \theta, \cos \theta, 0) \cdot (-\sin \theta, 0, 0) \, d\theta &= \int_{0}^{\pi} \cos^2 \theta + \sin^2 \theta \, d\theta \\ &= \pi. \end{align} $$