1. ## Ellipses

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

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

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

2. There's certainly something wrong. Surely the answer should be $\displaystyle \tfrac12ab(\theta_2 - \theta_1)$ (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.

3. I just thought of another way of doing it:

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

In my original post I let $\displaystyle r=1$, 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?

4. 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 $\displaystyle (r\cos\theta,r\sin\theta)$, but to the point $\displaystyle \psi(r,\theta) = (ar\cos\theta,br\sin\theta)$.

When you transform an integral $\displaystyle \iint f(x,y)\,dxdy$ 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 $\displaystyle \begin{vmatrix}\frac{\partial}{\partial r}(ar\cos\theta)& \frac{\partial}{\partial \theta}(ar\cos\theta)\\ \frac{\partial}{\partial r}(br\sin\theta)& \frac{\partial}{\partial \theta}(br\sin\theta)\end{vmatrix} = \begin{vmatrix}a\cos\theta& -ar\sin\theta\\ b\sin\theta& br\cos\theta\end{vmatrix} = abr$.

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