Let $\mathbf{F}(x,y) = \frac{-y \ \mathbf{i} + x \ \mathbf{j}}{\sqrt{x^2 + y^2}}$. Compute $\int_C \mathbf{F} \cdot d\mathbf{r}$ where $C$ is the circle of radius $r$, centered at the origin and traversed counterclockwise.
A natural parameterization of the circular path is given by the angle $\theta$. Because $C$ is a circle centered at the origin, the term $\sqrt{x^2 + y^2} = r$ is a constant along $C$. Hence, the integrand becomes fairly simple, and we choose direct parameterization as our method of integration.
Because the curve is a circle, we parameterize it with the angle $\theta$. Thus, we need an expression relating the $x$ and $y$ coordinates of each point on the circle to $\theta$:
Then, $$ \mathbf{r}(\theta) = r \cos \theta \ \mathbf{i} + r \sin \theta \ \mathbf{j}.$$
Traversing the circle counterclockwise means $\theta$ ranges from $0$ to $2 \pi$.
For convenience, we sketch the curve $C$ and its parameterization:
Step 2 - Express everything in terms of $\theta$
Now we express the integral exclusively in terms of $\theta$.
Expressing $d\mathbf{r}$ in terms of $\theta$: $$ d\mathbf{r}(\theta) = (- r \sin \theta \ \mathbf{i} + r \cos \theta \ \mathbf{j}) d\theta $$
Expressing $\mathbf{F} \cdot d\mathbf{r}$ in terms of $\theta$:$$\mathbf{F} \cdot d\mathbf{r} = (r \sin^2(\theta) + r \cos^2(\theta)) d\theta = r d\theta$$
Noting that the bounds of integration are from $\theta = 0$ to $\theta = 2\pi$, we write the integral as: $$ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_0^{2 \pi} r d\theta$$
Step 3 - Evaluate one-dimensional integral
The integral in $\theta$ evaluates to $2 \pi r$. Hence, $$ \int_C \mathbf{F}\cdot d\mathbf{r} = 2 \pi r.$$
