Finding the area of an ellipse using Divergence Theorem As an example of Divergence Theorem, our textbook mentions finding area of an ellipse, but it isn't clear how it was derived though.  
Following is an excerpt from the textbook.
Suppose there is an ellipse with the following equation,
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
Then we could parameterize it into
$$x=a\cos t, y=b\sin t \space (0\le t\le 2\pi)$$
In order to find the area enclosed by the ellipse, we use the equation $$area=\frac 12\int\mathbf r\cdot \mathbf n \space ds$$
where $$\mathbf r(x,y)=x\mathbf i+y\mathbf j$$
$$\mathbf n=(b\cos t,a\sin t)/v(t)\space (v(t)=\sqrt{b^2\cos^2t+a^2\sin^2t})$$
Hence, the area of the ellipse is
$$\begin{align} 
area & = \frac 12\int\mathbf r\cdot \mathbf n \space ds \\
& = \frac 12 \int_0^{2\pi} (a\cos t, b\sin t)\cdot(b\cos t, a\sin t)dt \\
& = ab\pi
\end{align}$$
What I'm not following is that how we could just ignore $v(t)$ when substituting $\mathbf n$? Or have I missed something along the way?
EDIT: I've actually tried throwing the entire equation (including $v(t)$) into wolframalpha, but, unfortunately, the engine couldn't return any meaning results (calculation timeout).
 A: Under a little bit more general settings, a line integral of a vector field $(p,q)$ along some oriented planar curve $L$ has the following form.
$$I=\int_L\left(pdx+qdy\right).$$ 
Denote $\mathbf{r}=(x,y)$ and let $L$ be parameterized as $t\mapsto\mathbf{r}(t)$, $t\in[0,1]$. We can always choose another parameter $s$, the so called arc length parameter, which satisfies $\frac{d\mathbf{r}}{ds}\equiv 1$, or $ds=|\mathbf{r}'(t)|dt$. Follow your notations, $\frac{ds}{dt}(t)=v(t)$. The unit normal vector field $\mathbf{n}$ can be represented as 
$$\mathbf{n}=(\frac{dy}{ds},-\frac{dx}{ds})=v(t)^{-1}(\frac{dy}{dt},-\frac{dx}{dt}).$$ 
If we choose $\mathbf{f}=(q,-p)$ as a vector field, then the line integral above can be rewritten as $$I=\int\mathbf{f}\cdot\mathbf{n}ds=\int\mathbf{f}\cdot(\frac{dy}{dt},-\frac{dx}{dt})dt.$$
Now we know that  when $L$ is a simple closed curve, and if we choose $\mathbf{f}=\frac{1}{2}\mathbf{r}$, the line integral above gives the area bounded by $L$, i.e. 
$$area=\frac{1}{2}\int\mathbf{r}\cdot\mathbf{n}ds=\frac{1}{2}\int(x\frac{dy}{dt}-y\frac{dx}{dt})dt=\frac{1}{2}\int_L( xdy-ydx).$$
