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战巡
Posted at 2013-10-14 06:23:34
回复 1# 青青子衿
没必要这么暴力吧...
图形是可以旋转的嘛...
整个东西逆时针转45度，圆变成$x^2+(y-\sqrt{2})^2=1$，双曲线变成$y^2-x^2=2$
这个方程组就好解多了嘛，可以得到
\[x_1=-\sqrt{-\frac{1}{2}+\sqrt{2}}, x_2=\sqrt{-\frac{1}{2}+\sqrt{2}}\]
面积为
\[\int_{-\sqrt{-\frac{1}{2}+\sqrt{2}}}^{\sqrt{-\frac{1}{2}+\sqrt{2}}}[\sqrt{1-x^2}+\sqrt{2}-\sqrt{2+x^2}]dx=\sqrt{2\sqrt{2}-1}+arcsin(\sqrt{-\frac{1}{2}+\sqrt{2}})-2arcsh(\frac{\sqrt{2\sqrt{2}-1}}{2})\] |
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