Matlab：闭合曲线和直线之间的面积

计算闭合曲线和直线之间的上面积的最佳方法是什么

alpha = [0:pi/360:2*pi];
x = 2.*(1-cos(alpha)).*cos(alpha);
y = 2.*(1-cos(alpha)).*sin(alpha);
w = [-4:0.1:0.5];
z = 0.5*w;
plot(x,y);
hold on;
grid on;
plot(w,z);

即使它是一条闭合曲线，同时占据正y轴和负y轴，也可以使用trapz来解决它吗？我看到了很多关于闭合曲线或曲线和直线的话题。但没有一条是关于曲线和直线的

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谢谢。

您只需使用interp1计算所需范围的面积即可。首先计算曲线和x轴之间的面积（超过）：

然后，直线和x轴（下方）之间的区域（x轴从交点到零）：

x3 = [-3.38 0];
r1 = interp1(x2,y2,x3);
a2 = abs(trapz(x3,r1));

然后对x轴下的曲线的其余部分做同样的处理：

x4 = [-3.99 -3.38];
r2 = interp1(x,y,x4);
a3 = abs(trapz(x4,r2))/2;

总面积为：

at = a1 + a2 + a3;

这里是整个代码：

clear all;
clc;

z = 0:0.01:2*pi;

x = 2.*(1-cos(z)).*cos(z);
y = 2.*(1-cos(z)).*sin(z);

plot(x,y);
hold on;
grid on;

x2 = -4:0.01:0.5;
y2 = 0.5*x2;

plot(x2,y2);

a1 = abs(trapz(x,y)/2);

x3 = [-3.38 0];
r1 = interp1(x2,y2,x3);
a2 = abs(trapz(x3,r1));

x4 = [-3.99 -3.38];
r2 = interp1(x,y,x4);
a3 = abs(trapz(x4,r2))/2;

at = a1 + a2 + a3;

clear all;
clc;

z = 0:0.01:2*pi;

x = 2.*(1-cos(z)).*cos(z);
y = 2.*(1-cos(z)).*sin(z);

plot(x,y);
hold on;
grid on;

x2 = -4:0.01:0.5;
y2 = 0.5*x2;

plot(x2,y2);

a1 = abs(trapz(x,y)/2);

x3 = [-3.38 0];
r1 = interp1(x2,y2,x3);
a2 = abs(trapz(x3,r1));

x4 = [-3.99 -3.38];
r2 = interp1(x,y,x4);
a3 = abs(trapz(x4,r2))/2;

at = a1 + a2 + a3;