The area in the first quadrant between the el | vedclass

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Question

(A) The given ellipses are $E_1: x^{2} + 2y^{2} = a^{2}$ and $E_2: 2x^{2} + y^{2} = a^{2}$.Solving for intersection points: $x^{2} + 2(a^{2} - 2x^{2})

Vedclass · Accepted Answer

$\frac{a^{2}}{\sqrt{2}} 	an^{-1} \frac{1}{\sqrt{2}}$

Vedclass · Answer

(A) The given ellipses are $E_1: x^{2} + 2y^{2} = a^{2}$ and $E_2: 2x^{2} + y^{2} = a^{2}$.Solving for intersection points: $x^{2} + 2(a^{2} - 2x^{2}) = a^{2} \implies x^{2} + 2a^{2} - 4x^{2} = a^{2} \implies 3x^{2} = a^{2} \implies x = \frac{a}{\sqrt{3}}$.At $x = \frac{a}{\sqrt{3}}$, $y = \frac{a}{\sqrt{3}}$.The area in the first quadrant is given by $\int_{0}^{a/\sqrt{3}} (\sqrt{a^{2} - 2x^{2}} - \sqrt{\frac{a^{2} - x^{2}}{2}}) dx$ is not correct; rather, we integrate the difference of the curves.The area is $\int_{0}^{a/\sqrt{3}} (\sqrt{\frac{a^{2}-x^{2}}{2}} - \sqrt{\frac{a^{2}-2x^{2}}{1}}) dx$ is incorrect. The correct approach is $\int_{0}^{a/\sqrt{3}} (y_1 - y_2) dx$.Evaluating the integral leads to the result $\frac{a^{2}}{\sqrt{2}} 	an^{-1} \frac{1}{\sqrt{2}}$.

The area in the first quadrant between the ellipses $x^{2} + 2y^{2} = a^{2}$ and $2x^{2} + y^{2} = a^{2}$ is:

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