# 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).

-
$v(t)$ has not been not ignored. Note that $ds=v(t)dt$. Maybe an alternative formulation of the line integral is less confusing for you: area=$\frac{1}{2}\int( xd y-ydx)$. – 23rd Nov 6 '12 at 12:27
@richard: You could post that as an answer. – joriki Nov 6 '12 at 12:28
@joriki: I am not quite familiar with the routine of this website. I saw lots of people give answers as comments rather than post a formal answer. Is this not encouraged? – 23rd Nov 6 '12 at 12:46
@richard: Yes, unfortunately people often do that, but if the comment actually answers the question and there isn't much more to say, the effect is often that that's the end of it and no-one ever posts an answer and the question remains unanswered and will regularly be bumped onto the main page, taking up valuable space there. So it depends on whether you're answering the question. In the present case it seems that your explanation that $v(t)$ hasn't been ignored clears things up and there will probably not be much more to say. – joriki Nov 6 '12 at 14:23
@joriki:Thank you for your patient explanation. I will post an answer. – 23rd Nov 6 '12 at 14:58

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. $$area=\frac{1}{2}\int\mathbf{r}\cdot\mathbf{n}ds=\frac{1}{2}\int(x\frac{dy}{dt}-y\frac{dx}{dt})dt=\frac{1}{2}\int_L( xdy-ydx).$$