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I'm trying to find the area inside the intersection of the region bounded by the ellipses $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ using vector calculus methods.

I understand how to calculate area of regions by taking the line integral around the boundary and using Green's Theorem. However what I'm stuck with is finding a paramterization of the boundary! Can anyone point me in the right direction?

Ng Chung Tak
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jonathan
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    You can always parameterize an ellipses by $x=a\cos t$ and $y=b\sin t$. Find just one of the intersecting point to know where you should switch to the other ellipse. – Hamid Enki Jan 09 '18 at 21:25
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    There’s no need to “switch to the other ellipse.” Take advantage of symmetry. – amd Jan 09 '18 at 21:30
  • @DougM ok, but which ellipse am I integrating? I don't know where each ellipse is, if you get me – jonathan Jan 09 '18 at 21:50

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I suppose $a<b$. The intersection points of the two ellipses are (see the figure where $a=2$ and $b=4$) $A,B,C,D =(\pm k,\pm k)$ with $$k=\frac{ab}{\sqrt{a^2+b^2}}$$

enter image description here

and, by symmetry, the searched area is $$ 4\left(k^2+\frac{2b}{a}\int_k^a \sqrt{a^2-x^2}dx\right) $$

Emilio Novati
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The figure will be symmetric across the line $x = y$ (along with $x = 0$ and $y = 0$)

We can analyze $\frac 18$ the figure, and then use this symmetry to our advantage.

Suppose $a<b$

$\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$ will be on the "inside" from $0$ to the point of intersection.

Parameterize the area inside the curve:

$x = ar\cos\theta\\ y = br\sin\theta\\ dx\ dy = ab r\ dr\ d\theta$

The last line is the Jacobean for this coordinate system.

And the point of intersection is where $x = y$

$a\cos\theta = b \sin \theta\\ \tan \theta = \frac {a}{b}$

$8\int_0^{\arctan \frac {a}{b}}\int_0^{1} abr\ dr\ d\theta$

$4 ab \arctan\frac {a}{b}$

Doug M
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Consider the picture below:

enter image description here

The description of two ellipses in polar coordinates is: $$r^2={1\over {{\cos^2\theta\over{a^2}}+{\sin^2\theta\over{b^2}}}}$$ $$r^2={1\over {{\cos^2\theta\over{b^2}}+{\sin^2\theta\over{a^2}}}}$$ so the surface we seek to find (assuming $a>b$) is: $$S=\int_{\pi\over4}^{3\pi\over4}{r^2\over2}d\theta={1\over 2}\int_{\pi\over4}^{3\pi\over4}{1\over {{\cos^2\theta\over{a^2}}+{\sin^2\theta\over{b^2}}}}d\theta=abtan^{-1}{a\over b}$$

Then the area between two ellipses is $\Large 4abtan^{-1}{max(a,b)\over min(a,b)}$

Mostafa Ayaz
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  • hi, thanks for this! can you explain your reasoning behind using that integrand? are not integrating only one ellipse here? – jonathan Jan 09 '18 at 21:31
  • to be specific, I can't seem to be able to get an image for this in my head – jonathan Jan 09 '18 at 21:38
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    Note that $$\tan^{-1} \frac{a}{b}+\tan^{-1} \frac{b}{a}=\frac{\pi}{2}$$ and the red region should be less than one quarter of the whole ellipse. So, the area of the red region should read $$ab\tan^{-1} \frac{b}{a}<\frac{\pi ab}{4}$$ Please double check your work. – Ng Chung Tak Jan 31 '18 at 11:50