Check Stokes' theorem using the function
\[\mathbf{v}=ay\hat{\mathbf{x}} +bx\hat{\mathbf{y}} \](where $a $ and $b $ are constants) and the circular path of radius $R $, centered at the origin in the $xy $ plane.
Let's recall The Fundamental Theorem for curls, also known as Stokes' theorem:
\begin{equation} \int _{\mathcal{S}}(\nabla \times \mathbf{v})\cdot d\mathbf{a}=\oint _{\mathcal{P}}\mathbf{v}\cdot d\mathbf{l}\tag{Eq. 1.57, page 33} \end{equation}Given the circular path mentioned above, the left hand side's surface integral will be over the surface of the circle of radius $R $, while the line integral will be over its circumference.
The field to be integrated over the entire area is the curl of $\mathbf{v} $:
\begin{equation} \nabla \times \mathbf{v} = (-a+b) \hat{\mathbf{z}} \end{equation}This is a constant vector pointing perpendicularly in or out (depending on the vaulues of $a $ and $b $) our of the circle's surface.
The line integral will be withr respect to the $\mathbf{v} $ field itself. Here's a live demonstration of the $\mathbf{v} $ field. Feel free to use the sliders to modify the field.
Let's now calculate both integrals and compare the results.
Starting with the surface integral:
\begin{align*} \int _{\mathcal{S}} (\nabla \times \mathbf{v})\cdot d\mathbf{a} &= \int _{\mathcal{S}}(-a+b)\hat{\mathbf{z}}\cdot da \hat{\mathbf{z}} \\ &= \int _{\mathcal{S}}(-a+b)da \\ &= (-a+b)\int _{\mathcal{S}}da\\ &= (-a+b)\cdot \left( \text{area of a circle of radius $R $} \right) \\ &= (-a+b)\pi R^2 \end{align*}It is true that $d\mathbf{a}=dxdy\hat{\mathbf{z}} $, but keeping it ``symbolic'' can be helpful (in this particular problem) to show that the integral is just the are of the surface. Longer ways to solve this integral include actually writing $dxdy $ and writing the limits of integration to bound the area to a circle of radius $R $, or converting the area element into a polar area element (which then simplifies the integration limits).
Let's now do the line integral. A better way to think about line integrals is to think about the following expression:
\[\int _{\mathcal{C}}\mathbf{v}\cdot d\mathbf{r} \]Where $\mathcal{C} $ is a Curve, a fancy word for a path. This curve depends on a single parameter which we usually denote with $t $. You can think of a curve as a vector that depends on a single variable.
Notice that a circle of radius $R $ can be represented by the following curve:
\[\text{Circle of radius $R $} =\left\lbrace \begin{pmatrix} R\cos \left( t \right) \\ R\sin \left( t \right) \\ \end{pmatrix} |t\in \left[ 0,2\pi \right] \right\rbrace \]That is, we will start to only deal with the variable $t $ rather than $x $ and $y $ itself. Using this variable, we see that the vector field $\mathbf{v} $ can be rewritten as:
\[\mathbf{v}=\begin{pmatrix} ay \\ bx \\ \end{pmatrix} \rightarrow \begin{pmatrix} a R \sin \left( t \right) \\ b R \cos \left( t \right) \\ \end{pmatrix} \]Notice then that for a given $t $, we now know how to find the Cartesian point of the curve AND the components of the vector field at that point on the curve:
And we have the infinitesimal:
\begin{align*} d\mathbf{r} &= \frac{d\mathbf{r}}{dt}dt \\ &= \begin{pmatrix} -R\sin \left( t \right) \\ R\cos \left( t \right) \\ \end{pmatrix} dt \end{align*}So the line integral is:
\begin{align*} \int \mathbf{v}\cdot d\mathbf{r} &= \int_{0}^{2\pi } \begin{pmatrix} a R \sin \left( t \right) \\ b R \cos \left( t \right) \\ \end{pmatrix}\cdot \begin{pmatrix} -R\sin \left( t \right) \\ R\cos \left( t \right) \\ \end{pmatrix} dt \\ &= \int_{0}^{2\pi } -aR^2 \sin ^2 (t) +b R ^2 \cos ^2 (t) dt\\ &= \int_{0}^{2\pi } -aR^2 \sin ^2 (t) dt+\int_{0}^{2\pi }b R ^2 \cos ^2 (t) dt \\ &= -a R^2 \int_{0}^{2\pi }\sin ^2 t dt + b R^2 \int_{0}^{2\pi }\cos ^2 (t)dt\\ &= -a R^2 \left( \pi \right) + b R^2 \left( \pi \right) \\ &= (-a+b)\pi R^2 \end{align*}Which is precisely the result we obtained by performing the surface integral of the curl of the field.