2016 AMC 10B 第 21 题

Question

方程

x^2+y^2=|x|+|y|

的图形围成区域的面积是多少？

What is the area of the region enclosed by the graph of the equation

x^2+y^2=|x|+|y|?

\ \pi+\sqrt{2}

\ \pi+2

\ \pi+2\sqrt{2}

\ 2\pi+\sqrt{2}

\ 2\pi+2\sqrt{2}

先在第一象限求面积，再乘以

4

。

Work in one quadrant and multiply by

4

第一象限的区域是一个三角形加一个半圆。

The first-quadrant region is a triangle plus a semicircle

这个方程在四个象限中都是对称的。在第一象限里它化为

x^2+y^2=x+y\text{，}

即

(x-\tfrac12)^2+(y-\tfrac12)^2=\tfrac12\text{。}

在第一象限中，围成的区域是直线

x+y=1

下方的三角形，面积为

\frac12

，再加上一个半径为

\sqrt{\frac{1}{2}}

的半圆，面积为

\frac\pi4

。

乘以

4

，总面积为

4\left(\frac12+\frac\pi4\right)=2+\pi\text{。}

所以正确答案是 B。

The equation is symmetric in all four quadrants. In the first quadrant it becomes

x^2+y^2=x+y,

or

(x-\tfrac12)^2+(y-\tfrac12)^2=\tfrac12.

In the first quadrant, the enclosed region is the triangle under

x+y=1

, with area

\frac12

, plus a semicircle of radius

\sqrt{\frac{1}{2}}

, with area

\frac\pi4

.

Multiplying by

4

, the total area is

4\left(\frac12+\frac\pi4\right)=2+\pi.

Thus, the correct answer is B.

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