MATLAB：闭曲线与直线之间的面积

Question

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);

Answer 1

您只需要使用interp1来计算所需范围的面积。首先，计算曲线和x轴之间的面积（over）：

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

然后，从交点到零时，直线和x轴之间的面积（下）：

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

然后在x轴下的曲线的其余部分也是这样：

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;