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For the two ellipses x²/a² + y²/b² = 1 and x²/b² + y²/a² = 1, where 0<b<a, what is the area of their overlapping region?

1. (a+b)²tan⁻¹(b/a)
2. (a+b)²tan⁻¹(a/b)
3. 4abtan⁻¹(b/a)
4. 4abtan⁻¹(a/b)

Correct answer: 4abtan⁻¹(b/a)

Solution

The two ellipses intersect on the lines y=+-x. Integrating the common interior region gives the standard result area = 4ab*arctan(b/a); numerical integration for a=2,b=1 gives 3.709, matching 4ab*arctan(b/a)=3.709.

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