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Last edited by hbghlyj 2025-5-6 21:48圆$A$的圆心$A$为双曲线$xy=1$一分支一点，使圆$A$与$x$，$y$轴都相切，求阴影部分的面积。
（1）求积分区间
设方程 $x y=1$ 与方程 $(x-1)^2+(y-1)^2=1$ 交于 $B, C$ 两点，则
\[
\begin{aligned}
& \left\{\begin{array} { l }
{ x y = 1 } \\
{ ( x - 1 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
y=\frac{1}{x} \\
x^2+y^2-2 x-2 y+1=0
\end{array}\right.\right. \\
& \Rightarrow x^2+\frac{1}{x^2}-2 x-\frac{2}{x}+1=0 \\
& \text {令 } x+\frac{1}{x}=t \text {, 则 } x^2+\frac{1}{x^2}=t^2-2
\end{aligned}
\]
$\therefore$ 原方程等价于 $t^2-2 t-1=0$
\[
\begin{aligned}
& \therefore t=1 \pm \sqrt{2} \\
& \therefore x+\frac{1}{x}=1 \pm \sqrt{2} \Rightarrow x^2-(1+\sqrt{2}) x+1=0 \text { 或 } x^2-(1-\sqrt{2}) x+1=0
\end{aligned}
\]
经检验 $x^2-(1-\sqrt{2}) x+1=0$ 无实根，舍去。
\[
\begin{aligned}
& \therefore x^2-(1+\sqrt{2}) x+1=0 \\
& \therefore x=\frac{1+\sqrt{2} \pm \sqrt{2 \sqrt{2}-1}}{2}, y=\frac{2}{1+\sqrt{2} \pm \sqrt{2 \sqrt{2}-1}} \\
& \therefore B\left(\frac{1+\sqrt{2}-\sqrt{2 \sqrt{2}-1}}{2}, \frac{2}{1+\sqrt{2}-\sqrt{2 \sqrt{2}-1}}\right), \\
& C\left(\frac{1+\sqrt{2}+\sqrt{2 \sqrt{2}-1}}{2}, \frac{2}{1+\sqrt{2}+\sqrt{2 \sqrt{2}-1}}\right)
\end{aligned}
\]
（2）定面积元素
\[
\begin{aligned}
& d A_1=\left[\sqrt{-x^2+2 x}-1-\frac{1}{x}\right] d x, x \in\left[\frac{1+\sqrt{2}-\sqrt{2 \sqrt{2}-1}}{2}, \frac{1+\sqrt{2}+\sqrt{2 \sqrt{2}-1}}{2}\right], \\
& d A_2=\left[\sqrt{-x^2+2 x}-1-\left(-\sqrt{-x^2+2 x}-1\right)\right] d x, x \in\left[\frac{1+\sqrt{2}+\sqrt{2 \sqrt{2}-1}}{2}, 2\right]
\end{aligned}
\]
（3）求面积
\[
A=\int_{\frac{1+\sqrt{2}-\sqrt{2 \sqrt{2}-1}}{2}}^{\frac{1+\sqrt{2}+\sqrt{\sqrt{2}-1}}{2}} d A_1+\int_{\frac{1+\sqrt{2}}{2}+\sqrt{2 \sqrt{2}-1}}^2 d A_2=?
\] |
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战巡
Posted 2013-10-14 06:23
