Ellipses

**Showcase_22** · April 23rd 2009, 11:30 AM

Given $0< a \leq b,$ define $\underline{\psi}:[0, 2 + \infty) \times [0, 2 \pi] \rightarrow \mathbb{R}^2$ by $\underline{\psi}(r, \theta):=(arcos(\theta),br sin(\theta)).$

Given $0 \leq \theta_1< \theta_2 \leq 2 \pi$ , calculate the area of the sector of the ellipse $\left( \frac{x^2}{a^2} \right)+\left( \frac{y^2}{b^2} \right)=1$ defined by $\{ \underline{\psi}(r, \theta)|0 \leq r \leq 1, \theta_1 \leq \theta \leq \theta_2 \}$ .

My answer is $\int_{\theta_1}^{\theta_2} \int_{0}^{1}dr \ d \theta=\theta_2- \theta_1$

Is this correct? I'm slightly worried about my r limits...

**Opalg** · April 23rd 2009, 01:18 PM

There's certainly something wrong. Surely the answer should be $\tfrac12ab(\theta_2 - \theta_1)$ (a and b must come into it somewhere). But at this time of the evening I don't immediately see how to set up the integral. Maybe my brain will be fresher tomorrow morning.

**Showcase_22** · April 23rd 2009, 01:26 PM

I just thought of another way of doing it:

$x=rcos(\theta)$ and $y=rsin(\theta)$ and then substitute these into the equation for the ellipse. This will give an expression for r in terms of $\theta_2$ and $\theta_1$ .

In my original post I let $r=1$ , but the integrand would now be the expression I calculated from above.

It's a little late to try this now, i'll try it tomorrow.

Does anyone think this method will work?

**Opalg** · April 23rd 2009, 11:40 PM

The difficulty with this problem is that the notation is misleading. The variables r and θ are not polar coordinates. They refer not to the point $(r\cos\theta,r\sin\theta)$ , but to the point $\psi(r,\theta) = (ar\cos\theta,br\sin\theta)$ .

When you transform an integral $\iint f(x,y)\,dxdy$ to polar coordinates, you have to replace dxdy by rdrdθ. The reason for the additional r is that it is the absolute value of the Jacobian determinant of the change of coordinates. For the variables r, θ in this problem, the Jacobian determinant is $\begin{vmatrix}\frac{\partial}{\partial r}(ar\cos\theta)& \frac{\partial}{\partial \theta}(ar\cos\theta)\\ \frac{\partial}{\partial r}(br\sin\theta)& \frac{\partial}{\partial \theta}(br\sin\theta)\end{vmatrix} = \begin{vmatrix}a\cos\theta& -ar\sin\theta\\ b\sin\theta& br\cos\theta\end{vmatrix} = abr$ .

So the integral for the area of that sector of the ellipse is $\int_{\theta_1}^{\theta_2}\!\!\int_0^1abr\,drd\the ta = \tfrac12ab(\theta_2-\theta_1)$ .

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