https://spherical-harmonics.hateblo.jp/entry/Todai/1933/Kougaku

2022.08.10記

[1] 原點ヲ含ミ，次ノ二曲線ニヨリ境サルヽ部分ノ面積ヲ計算セヨ．
$x^2+2y^2=2a^2$ ， $2(x^2-y^2)=a^2$

2022.08.11記

[解答]
2曲線の交点の座標は $\left(\pm a, \pm\dfrac{a}{\sqrt{2}}\right)$ (複号任意) であるから，求める面積は
$S=4\displaystyle\int_0^a \dfrac{1}{\sqrt{2}}\sqrt{2a^2-x^2}\,dx-4\displaystyle\int_{a/\sqrt{2}}^{a}\sqrt{x^2-\dfrac{a^2}{2}}\,dx$
$=\sqrt{2}\Bigl[ x\sqrt{2a^2-x^2}+2a^2\mbox{Arcsin}\,\dfrac{x}{\sqrt{2}a}\Bigr]_0^a$ $-2\Bigl[ x\sqrt{x^2-\dfrac{a^2}{2}}-\dfrac{a^2}{2}\log\left|x+\sqrt{x^2-\dfrac{a^2}{2}}\right|\Bigr]_{a/\sqrt{2}}^{a}$
$=\sqrt{2}\left( a\sqrt{a^2}+\dfrac{\pi a^2}{2}\right)$ $-2\left( a\sqrt{\dfrac{a^2}{2}}-\dfrac{a^2}{2}\log\left| a+\sqrt{\dfrac{a^2}{2}} \right|\right)$ $-a^2\log\left|\dfrac{a}{\sqrt{2}}\right|$
$=\dfrac{\pi a^2}{\sqrt{2}}$ $+a^2\log(1+\sqrt{2})$