(856,#11) Use a double integral to find the area of the region enclosed by the lemniscate r² = 4 cos(2 theta).

The geometry: The two loops, r = 2 sqrt(2 theta) and r = -2 sqrt(2 theta) are graphed below. The blue line is y = x, in polar coordinates, theta = Pi/4; the red is theta = -Pi/4

p1:= plot([2*sqrt(cos(2*t)), -2*sqrt(cos(2*t))], t = -Pi/4..Pi/4, coords = polar, color = [blue,red], tickmarks = [4,3],numpoints = 200):
p2 := plot([x, -x], x = -1.5..1.5, color = [blue,red]):
display({p1,p2}, scaling = constrained, view = [-2..2, -1..1]);

The limits of integration are theta = -Pi/4..Pi/4 and r = 0..2sqrt(2 theta). The value of the integral is 4, which is obtained by doubling the area of one loop:

f := (x,y) ->1:
c := t -> 0: d := t ->2*sqrt(cos(2*t)):
Integrate_rdrdt(f, -Pi/4,Pi/4,c,d);

echo:f(r*cos(t), r*sin(t))*r = r c(t) = 0 d(t) = 2*cos(2*t)^(1/2)

int(f(r*cos(t), r*sin(t))*r, r ))= 1/2*r^2

int(f(r*cos(t), r*sin(t))*r, r = c(t)..d(t))= 4*cos(t)^2-2

int(int(f(r*cos(t), r*sin(t))*r, r = c(t)..d(t)),t)= 2*cos(t)*sin(t)

int(int(f(r*cos(t), r*sin(t))*r, r = c(t) ..d(t)), t = a..b) = 2

by:sh