Dead Polar Flower

With the function $r(\theta) = \theta(\theta-2)(\theta-4)(\theta-6)$ , we need to start by finding the real roots of the equation $r(\theta) + r(\theta+\pi) = 0$ Solutions of this equation give the polar angles of the two points where the curve self-intersects (apart from the origin). These roots are $\alpha \; = \; 0.1977262107572581 \hspace{2cm} \beta \; =\; 2.66068113565295$ We note that symmetry $r(\theta) = r(6-\theta)$ of the curve tells us that $\alpha + \beta + \pi = 6$ .

Drawing radial lines to these points of self-intersection, the required area is $\begin{array}{rcl} A & = & \displaystyle \int_\alpha^2 \tfrac12r(\theta)^2\,d\theta + \int_\beta^{\alpha+\pi} \tfrac12 r(\theta)^2\,d\theta + \int_4^{\beta+\pi} \tfrac12r(\theta)^2\,d\theta \\ & = & \displaystyle \int_\alpha^2 r(\theta)^2\,d\theta + \int_\beta^3 r(\theta)^2\,d\theta \; =\; \boxed{275.121702565}\end{array}$