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Math Help - Ellipses

  1. #1
    Super Member Showcase_22's Avatar
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    Ellipses

    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...
    Last edited by Showcase_22; April 24th 2009 at 01:41 AM.
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  2. #2
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    Opalg's Avatar
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    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.
    Last edited by Opalg; April 24th 2009 at 12:41 AM. Reason: Forgot the factor 1/2
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  3. #3
    Super Member Showcase_22's Avatar
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    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?
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  4. #4
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    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).
    Last edited by Opalg; April 24th 2009 at 11:47 AM.
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