---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- # Let's calculate the area of the circle sector depictedbelow on the leftBF=3; HF=3 PLANE(0,5,0,5) f=function(x) sqrt(25-x^2); g=function(x) 3/4*x graph1(f,-5,5,"blue"); graph1(g,-5,5,"blue") graph(f,4,5,"red"); graph(g,0,4,"red"); segm(0,0, 5,0, "red")#This is the figure. It is a circle segment. I can take the circle area multiplied by # the ratio between the angle of the segment and 360°:A=c(5,0); B=c(0,0); C=c(4,3); angle(A,B,C) # 36.8699 pi*5^2*angle(A,B,C)/360 # 8.043764 This is the searched area#Alternatively, I calculate the area of the triangle and that of the remaining part # where the figure is broken by the vertical segment falling from (4,3)polyl(c(4,4),c(0,3),"blue") # The area of the triangle plus the integral of f between 4 and 5: 4*3/2 + integral(f,4,5) # 8.043764# Let's calculate the area of the figureabove, in the center:PLANE(0,1, 0,1) f=function(x) x^2; g=function(x) x graph(f,0,1, "blue"); graph(g,0,1,"red")# The difference between the area under the graph of g and the one below that of f:integral(g,0,1)-integral(f,0,1) # 0.1666667 # Or: h = function(x) g(x)-h(x); integral(h,0,1) # 0.1666667 which is equivalent to: fraction(integral(h,0,1)) # 1/6 This is the searched area # # The area of the figureabove, on the right, between the two curves:PLANE(0,4,-2,2) f = function(x) sin(x); g = function(x) (x-2)^2 graph(f,0,4, "red"); graph(g,0,4, "blue") solution2(f,g,1,2) # 1.064761 solution2(f,g,2,3) # 2.67242 h = function(x) f(x)-g(x) integral(h, solution2(f,g,1,2), solution2(f,g,2,3)) # 1.002636 This is the searched area # just over a "square" # # How to calculate areas enclosed bypolar curves. # In the case of a circle of radius R I have that area A is pi*R^2. # Generalizing, if R=R(t): # If the angle t changes from H to K: # RR = function(t) R(t)^2; A = integral(RR, H,K)/2 # First example:cardioid. Numerically we have: R = function(t) 1-sin(t); PLANE(-2,2, -3,1); polar(R, 0,2*pi, "blue") areaPolar(R,0,2*pi,1000) # 4.712327 areaPolar(R,0,2*pi,2000) # 4.712373 areaPolar(R,0,2*pi,4000) # 4.712385 areaPolar(R,0,2*pi,8000) # 4.712388 areaPolar(R,0,2*pi,16000) # 4.712389 # With integration: R=function(t) 1-sin(t); RR=function(t) R(t)^2; A=integral(RR, 0,2*pi)/2; A # 4.712389 A/pi # 1.5 A =3/2*pi# # Second example: thefigure above right. PLANE(-1.25,1.25, -1,1.5) R1 = function(t) sqrt(2)*sin(t); R2 = function(t) sqrt(sin(2*t)) polar(R1,0,2*pi, "seagreen"); polar(R2,0,2*pi, "brown") # I give color to the figure: P=function(x,y) {r=sqrt(x^2+y^2); t=dirArrow1(0,0, x,y)*pi/180; r<=sqrt(sin(t*2)) & r<=sqrt(2)*sin(t)} for(i in 1:8) FIGURE(P,0,1,0,1, "orange") polar(R1,0,2*pi, "seagreen"); polar(R2,0,2*pi, "brown") g = function(x) x; graph1(g, 0,2, "red") polar(R1,0,pi/4, "blue"); polar(R2,pi/4,pi/2, "red") R = function(t) ifelse(t < pi/4,R1(t),R2(t)); RR=function(t) R(t)^2; A=integral(RR,0,pi/2)/2; A # 0.3926991 A/pi # 0.25pi/8# 0.3926991 # # Another example: thelazy eight(ρ = √|cos(2θ|): # ( in the right figure θ is in [-π/4,π/4] and in [π-π/4,π+π/4] ) R = function(t) sqrt(abs(cos(2*t))); PLANE(-1.5,1.5, -1.5,1.5); polar(R, 0,2*pi, "blue") RR=function(t) R(t)^2; A=integral(RR,0,2*pi)/2; A #2